Syntax error unexpected string content expecting or identifier or variable or number - Исправление ошибок и поиск оптимальных решений проблем

Warning: Cannot modify header information — headers already sent

Happens when your script tries to send an HTTP header to the client but there already was output before, which resulted in headers to be already sent to the client.

This is an E_WARNING and it will not stop the script.

A typical example would be a template file like this:

<html>
    <?php session_start(); ?>
    <head><title>My Page</title>
</html>
...

The session_start() function will try to send headers with the session cookie to the client. But PHP already sent headers when it wrote the <html> element to the output stream. You’d have to move the session_start() to the top.

You can solve this by going through the lines before the code triggering the Warning and check where it outputs. Move any header sending code before that code.

An often overlooked output is new lines after PHP’s closing ?>. It is considered a standard practice to omit ?> when it is the last thing in the file. Likewise, another common cause for this warning is when the opening <?php has an empty space, line, or invisible character before it, causing the web server to send the headers and the whitespace/newline thus when PHP starts parsing won’t be able to submit any header.

If your file has more than one <?php ... ?> code block in it, you should not have any spaces in between them. (Note: You might have multiple blocks if you had code that was automatically constructed)

Also make sure you don’t have any Byte Order Marks in your code, for example when the encoding of the script is UTF-8 with BOM.

Related Questions:

Headers already sent by PHP
All PHP «Headers already sent» Questions on Stackoverflow
Byte Order Mark
What PHP Functions Create Output?

Fatal error: Call to a member function … on a non-object

Happens with code similar to xyz->method() where xyz is not an object and therefore that method can not be called.

This is a fatal error which will stop the script (forward compatibility notice: It will become a catchable error starting with PHP 7).

Most often this is a sign that the code has missing checks for error conditions. Validate that an object is actually an object before calling its methods.

A typical example would be

// ... some code using PDO
$statement = $pdo->prepare('invalid query', ...);
$statement->execute(...);

In the example above, the query cannot be prepared and prepare() will assign false to $statement. Trying to call the execute() method will then result in the Fatal Error because false is a «non-object» because the value is a boolean.

Figure out why your function returned a boolean instead of an object. For example, check the $pdo object for the last error that occurred. Details on how to debug this will depend on how errors are handled for the particular function/object/class in question.

If even the ->prepare is failing then your $pdo database handle object didn’t get passed into the current scope. Find where it got defined. Then pass it as a parameter, store it as property, or share it via the global scope.

Another problem may be conditionally creating an object and then trying to call a method outside that conditional block. For example

if ($someCondition) {
    $myObj = new MyObj();
}
// ...
$myObj->someMethod();

By attempting to execute the method outside the conditional block, your object may not be defined.

Related Questions:

Call to a member function on a non-object
List all PHP «Fatal error: Call to a member function … on a non-object» Questions on Stackoverflow

«Notice: Undefined Index», or «Warning: Undefined array key»

Happens when you try to access an array by a key that does not exist in the array.

A typical example of an Undefined Index notice would be (demo)

$data = array('foo' => '42', 'bar');
echo $data['spinach'];
echo $data[1];

Both spinach and 1 do not exist in the array, causing an E_NOTICE to be triggered. In PHP 8.0, this is an E_WARNING instead.

The solution is to make sure the index or offset exists prior to accessing that index. This may mean that you need to fix a bug in your program to ensure that those indexes do exist when you expect them to. Or it may mean that you need to test whether the indexes exist using array_key_exists or isset:

$data = array('foo' => '42', 'bar');
if (array_key_exists('spinach', $data)) {
    echo $data['spinach'];
}
else {
    echo 'No key spinach in the array';
}

If you have code like:

<?php echo $_POST['message']; ?>
<form method="post" action="">
    <input type="text" name="message">
    ...

then $_POST['message'] will not be set when this page is first loaded and you will get the above error. Only when the form is submitted and this code is run a second time will the array index exist. You typically check for this with:

if ($_POST)  ..  // if the $_POST array is not empty
// or
if ($_SERVER['REQUEST_METHOD'] == 'POST') ..  // page was requested with POST

Related Questions:

Reference: “Notice: Undefined variable” and “Notice: Undefined index”
All PHP «Notice: Undefined Index» Questions on Stackoverflow
http://php.net/arrays

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given

First and foremost:

Please, don’t use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi — this article will help you decide which. If you choose PDO, here is a good tutorial.

This happens when you try to fetch data from the result of mysql_query but the query failed.

This is a warning and won’t stop the script, but will make your program wrong.

You need to check the result returned by mysql_query by

$res = mysql_query($sql);
if (!$res) {
   trigger_error(mysql_error(),E_USER_ERROR);
}
// after checking, do the fetch

Related Questions:

mysql_fetch_array() expects parameter 1 to be resource, boolean given in select
All «mysql_fetch_array() expects parameter 1 to be resource, boolean given» Questions on Stackoverflow

Related Errors:

Warning: [function] expects parameter 1 to be resource, boolean given

Other mysql* functions that also expect a MySQL result resource as a parameter will produce the same error for the same reason.

Fatal error: Using $this when not in object context

$this is a special variable in PHP which can not be assigned. If it is accessed in a context where it does not exist, this fatal error is given.

This error can occur:

If a non-static method is called statically. Example:
```
class Foo {
   protected $var;
   public function __construct($var) {
       $this->var = $var;
   }

   public static function bar () {
       // ^^^^^^
       echo $this->var;
       //   ^^^^^
   }
}

Foo::bar();
```
How to fix: review your code again, $this can only be used in an object context, and should never be used in a static method. Also, a static method should not access the non-static property. Use self::$static_property to access the static property.
If code from a class method has been copied over into a normal function or just the global scope and keeping the $this special variable.
How to fix: Review the code and replace $this with a different substitution variable.

Related Questions:

Call non-static method as static: PHP Fatal error: Using $this when not in object context
Copy over code: Fatal error: Using $this when not in object context
All «Using $this when not in object context» Questions on Stackoverflow

Fatal error: Call to undefined function XXX

Happens when you try to call a function that is not defined yet. Common causes include missing extensions and includes, conditional function declaration, function in a function declaration or simple typos.

Example 1 — Conditional Function Declaration

$someCondition = false;
if ($someCondition === true) {
    function fn() {
        return 1;
    }
}
echo fn(); // triggers error

In this case, fn() will never be declared because $someCondition is not true.

Example 2 — Function in Function Declaration

function createFn() 
{
    function fn() {
        return 1;
    }
}
echo fn(); // triggers error

In this case, fn will only be declared once createFn() gets called. Note that subsequent calls to createFn() will trigger an error about Redeclaration of an Existing function.

You may also see this for a PHP built-in function. Try searching for the function in the official manual, and check what «extension» (PHP module) it belongs to, and what versions of PHP support it.

In case of a missing extension, install that extension and enable it in php.ini. Refer to the Installation Instructions in the PHP Manual for the extension your function appears in. You may also be able to enable or install the extension using your package manager (e.g. apt in Debian or Ubuntu, yum in Red Hat or CentOS), or a control panel in a shared hosting environment.

If the function was introduced in a newer version of PHP from what you are using, you may find links to alternative implementations in the manual or its comment section. If it has been removed from PHP, look for information about why, as it may no longer be necessary.

In case of missing includes, make sure to include the file declaring the function before calling the function.

In case of typos, fix the typo.

Related Questions:

https://stackoverflow.com/search?q=Fatal+error%3A+Call+to+undefined+function

Fatal error: Can’t use function return value in write context

This usually happens when using a function directly with empty.

Example:

if (empty(is_null(null))) {
  echo 'empty';
}

This is because empty is a language construct and not a function, it cannot be called with an expression as its argument in PHP versions before 5.5. Prior to PHP 5.5, the argument to empty() must be a variable, but an arbitrary expression (such as a return value of a function) is permissible in PHP 5.5+.

empty, despite its name, does not actually check if a variable is «empty». Instead, it checks if a variable doesn’t exist, or == false. Expressions (like is_null(null) in the example) will always be deemed to exist, so here empty is only checking if it is equal to false. You could replace empty() here with !, e.g. if (!is_null(null)), or explicitly compare to false, e.g. if (is_null(null) == false).

Related Questions:

Fatal error: Can’t use function the return value

MySQL: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near … at line …

This error is often caused because you forgot to properly escape the data passed to a MySQL query.

An example of what not to do (the «Bad Idea»):

$query = "UPDATE `posts` SET my_text='{$_POST['text']}' WHERE id={$_GET['id']}";
mysqli_query($db, $query);

This code could be included in a page with a form to submit, with an URL such as http://example.com/edit.php?id=10 (to edit the post n°10)

What will happen if the submitted text contains single quotes? $query will end up with:

$query = "UPDATE `posts` SET my_text='I'm a PHP newbie' WHERE id=10';

And when this query is sent to MySQL, it will complain that the syntax is wrong, because there is an extra single quote in the middle.

To avoid such errors, you MUST always escape the data before use in a query.

Escaping data before use in a SQL query is also very important because if you don’t, your script will be open to SQL injections. An SQL injection may cause alteration, loss or modification of a record, a table or an entire database. This is a very serious security issue!

Documentation:

How can I prevent SQL injection in PHP?
mysql_real_escape_string()
mysqli_real_escape_string()
How does the SQL injection from the «Bobby Tables» XKCD comic work?
SQL injection that gets around mysql_real_escape_string()

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE

In PHP 8.0 and above, the message is instead:

syntax error, unexpected string content «», expecting «-» or identifier or variable or number

This error is most often encountered when attempting to reference an array value with a quoted key for interpolation inside a double-quoted string when the entire complex variable construct is not enclosed in {}.

The error case:

This will result in Unexpected T_ENCAPSED_AND_WHITESPACE:

echo "This is a double-quoted string with a quoted array key in $array['key']";
//---------------------------------------------------------------------^^^^^

Possible fixes:

In a double-quoted string, PHP will permit array key strings to be used unquoted, and will not issue an E_NOTICE. So the above could be written as:

echo "This is a double-quoted string with an un-quoted array key in $array[key]";
//------------------------------------------------------------------------^^^^^

The entire complex array variable and key(s) can be enclosed in {}, in which case they should be quoted to avoid an E_NOTICE. The PHP documentation recommends this syntax for complex variables.

echo "This is a double-quoted string with a quoted array key in {$array['key']}";
//--------------------------------------------------------------^^^^^^^^^^^^^^^
// Or a complex array property of an object:
echo "This is a a double-quoted string with a complex {$object->property->array['key']}";

Of course, the alternative to any of the above is to concatenate the array variable in instead of interpolating it:

echo "This is a double-quoted string with an array variable". $array['key'] . " concatenated inside.";
//----------------------------------------------------------^^^^^^^^^^^^^^^^^^^^^

For reference, see the section on Variable Parsing in the PHP Strings manual page

Fatal error: Allowed memory size of XXX bytes exhausted (tried to allocate XXX bytes)

There is not enough memory to run your script. PHP has reached the memory limit and stops executing it. This error is fatal, the script stops. The value of the memory limit can be configured either in the php.ini file or by using ini_set('memory_limit', '128 M'); in the script (which will overwrite the value defined in php.ini). The purpose of the memory limit is to prevent a single PHP script from gobbling up all the available memory and bringing the whole web server down.

The first thing to do is to minimise the amount of memory your script needs. For instance, if you’re reading a large file into a variable or are fetching many records from a database and are storing them all in an array, that may use a lot of memory. Change your code to instead read the file line by line or fetch database records one at a time without storing them all in memory. This does require a bit of a conceptual awareness of what’s going on behind the scenes and when data is stored in memory vs. elsewhere.

If this error occurred when your script was not doing memory-intensive work, you need to check your code to see whether there is a memory leak. The memory_get_usage function is your friend.

Related Questions:

All «Fatal error: Allowed memory size of XXX bytes exhausted» Questions on Stackoverflow

Warning: [function]: failed to open stream: [reason]

It happens when you call a file usually by include, require or fopen and PHP couldn’t find the file or have not enough permission to load the file.

This can happen for a variety of reasons :

the file path is wrong
the file path is relative
include path is wrong
permissions are too restrictive
SELinux is in force
and many more …

One common mistake is to not use an absolute path. This can be easily solved by using a full path or magic constants like __DIR__ or dirname(__FILE__):

include __DIR__ . '/inc/globals.inc.php';

or:

require dirname(__FILE__) . '/inc/globals.inc.php';

Ensuring the right path is used is one step in troubleshooting these issues, this can also be related to non-existing files, rights of the filesystem preventing access or open basedir restrictions by PHP itself.

The best way to solve this problem quickly is to follow the troubleshooting checklist below.

Related Questions:

Troubleshooting checklist: Failed to open stream

Related Errors:

Warning: open_basedir restriction in effect

Notice: Undefined variable

Happens when you try to use a variable that wasn’t previously defined.

A typical example would be

foreach ($items as $item) {
    // do something with item
    $counter++;
}

If you didn’t define $counter before, the code above will trigger the notice.

The correct way is to set the variable before using it

$counter = 0;
foreach ($items as $item) {
    // do something with item
    $counter++;
}

Similarly, a variable is not accessible outside its scope, for example when using anonymous functions.

$prefix = "Blueberry";
$food = ["cake", "cheese", "pie"];
$prefixedFood = array_map(function ($food) {
  // Prefix is undefined
  return "${prefix} ${food}";
}, $food);

This should instead be passed using use

$prefix = "Blueberry";
$food = ["cake", "cheese", "pie"];
$prefixedFood = array_map(function ($food) use ($prefix) {
  return "${prefix} ${food}";
}, $food);

Notice: Undefined property

This error means much the same thing, but refers to a property of an object. Reusing the example above, this code would trigger the error because the counter property hasn’t been set.

$obj = new stdclass;
$obj->property = 2342;
foreach ($items as $item) {
    // do something with item
    $obj->counter++;
}

Related Questions:

All PHP «Notice: Undefined Variable» Questions on Stackoverflow
«Notice: Undefined variable», «Notice: Undefined index», and «Notice: Undefined offset» using PHP
Reference: What is variable scope, which variables are accessible from where and what are «undefined variable» errors?

Notice: Use of undefined constant XXX — assumed ‘XXX’

or, in PHP 7.2 or later:

Warning: Use of undefined constant XXX — assumed ‘XXX’ (this will throw an Error in a future version of PHP)

or, in PHP 8.0 or later:

Error: Undefined constant XXX

This occurs when a token is used in the code and appears to be a constant, but a constant by that name is not defined.

One of the most common causes of this notice is a failure to quote a string used as an associative array key.

For example:

// Wrong
echo $array[key];

// Right
echo $array['key'];

Another common cause is a missing $ (dollar) sign in front of a variable name:

// Wrong
echo varName;

// Right
echo $varName;

Or perhaps you have misspelled some other constant or keyword:

// Wrong
$foo = fasle;

// Right
$foo = false;

It can also be a sign that a needed PHP extension or library is missing when you try to access a constant defined by that library.

Related Questions:

What does the PHP error message “Notice: Use of undefined constant” mean?

Fatal error: Cannot redeclare class [class name]

Fatal error: Cannot redeclare [function name]

This means you’re either using the same function/class name twice and need to rename one of them, or it is because you have used require or include where you should be using require_once or include_once.

When a class or a function is declared in PHP, it is immutable, and cannot later be declared with a new value.

Consider the following code:

class.php

<?php

class MyClass
{
    public function doSomething()
    {
        // do stuff here
    }
}

index.php

<?php

function do_stuff()
{
   require 'class.php';
   $obj = new MyClass;
   $obj->doSomething();
}

do_stuff();
do_stuff();

The second call to do_stuff() will produce the error above. By changing require to require_once, we can be certain that the file that contains the definition of MyClass will only be loaded once, and the error will be avoided.

Parse error: syntax error, unexpected T_VARIABLE

Possible scenario

I can’t seem to find where my code has gone wrong. Here is my full error:

Parse error: syntax error, unexpected T_VARIABLE on line x

What I am trying

$sql = 'SELECT * FROM dealer WHERE id="'$id.'"';

Answer

Parse error: A problem with the syntax of your program, such as leaving a semicolon off of the end of a statement or, like the case above, missing the . operator. The interpreter stops running your program when it encounters a parse error.

In simple words this is a syntax error, meaning that there is something in your code stopping it from being parsed correctly and therefore running.

What you should do is check carefully at the lines around where the error is for any simple mistakes.

That error message means that in line x of the file, the PHP interpreter was expecting to see an open parenthesis but instead, it encountered something called T_VARIABLE. That T_VARIABLE thing is called a token. It’s the PHP interpreter’s way of expressing different fundamental parts of programs. When the interpreter reads in a program, it translates what you’ve written into a list of tokens. Wherever you put a variable in your program, there is aT_VARIABLE token in the interpreter’s list.

Good read: List of Parser Tokens

So make sure you enable at least E_PARSE in your php.ini. Parse errors should not exist in production scripts.

I always recommended to add the following statement, while coding:

error_reporting(E_ALL);

PHP error reporting

Also, a good idea to use an IDE which will let you know parse errors while typing. You can use:

NetBeans (a fine piece of beauty, free software) (the best in my opinion)
PhpStorm (uncle Gordon love this: P, paid plan, contains proprietary and free software)
Eclipse (beauty and the beast, free software)

Related Questions:

Reference: PHP Parse/Syntax Errors; and How to solve them?

Notice: Uninitialized string offset: `*`

As the name indicates, such type of error occurs, when you are most likely trying to iterate over or find a value from an array with a non-existing key.

Consider you, are trying to show every letter from $string

$string = 'ABCD'; 
for ($i=0, $len = strlen($string); $i <= $len; $i++){
    echo "$string[$i] n"; 
}

The above example will generate (online demo):

A
B
C
D
Notice: Uninitialized string offset: 4 in XXX on line X

And, as soon as the script finishes echoing D you’ll get the error, because inside the for() loop, you have told PHP to show you the from first to fifth string character from 'ABCD' Which, exists, but since the loop starts to count from 0 and echoes D by the time it reaches to 4, it will throw an offset error.

Similar Errors:

Illegal string offset ‘option 1’

Notice: Trying to get property of non-object error

Happens when you try to access a property of an object while there is no object.

A typical example for a non-object notice would be

$users = json_decode('[{"name": "hakre"}]');
echo $users->name; # Notice: Trying to get property of non-object

In this case, $users is an array (so not an object) and it does not have any properties.

This is similar to accessing a non-existing index or key of an array (see Notice: Undefined Index).

This example is much simplified. Most often such a notice signals an unchecked return value, e.g. when a library returns NULL if an object does not exists or just an unexpected non-object value (e.g. in an Xpath result, JSON structures with unexpected format, XML with unexpected format etc.) but the code does not check for such a condition.

As those non-objects are often processed further on, often a fatal-error happens next on calling an object method on a non-object (see: Fatal error: Call to a member function … on a non-object) halting the script.

It can be easily prevented by checking for error conditions and/or that a variable matches an expectation. Here such a notice with a DOMXPath example:

$result  = $xpath->query("//*[@id='detail-sections']/div[1]");
$divText = $result->item(0)->nodeValue; # Notice: Trying to get property of non-object

The problem is accessing the nodeValue property (field) of the first item while it has not been checked if it exists or not in the $result collection. Instead it pays to make the code more explicit by assigning variables to the objects the code operates on:

$result  = $xpath->query("//*[@id='detail-sections']/div[1]");
$div     = $result->item(0);
$divText = "-/-";
if (is_object($div)) {
    $divText = $div->nodeValue;
}
echo $divText;

Related errors:

Notice: Undefined Index
Fatal error: Call to a member function … on a non-object

Parse error: syntax error, unexpected ‘[‘

This error comes in two variatians:

Variation 1

$arr = [1, 2, 3];

This array initializer syntax was only introduced in PHP 5.4; it will raise a parser error on versions before that. If possible, upgrade your installation or use the old syntax:

$arr = array(1, 2, 3);

Variation 2

$suffix = explode(',', 'foo,bar')[1];

Array dereferencing function results was also introduced in PHP 5.4. If it’s not possible to upgrade you need to use a (temporary) variable:

$parts = explode(',', 'foo,bar');
$suffix = $parts[1];

Warning: [function] expects parameter 1 to be resource, boolean given

(A more general variation of Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given)

Resources are a type in PHP (like strings, integers or objects). A resource is an opaque blob with no inherently meaningful value of its own. A resource is specific to and defined by a certain set of PHP functions or extension. For instance, the Mysql extension defines two resource types:

There are two resource types used in the MySQL module. The first one is the link identifier for a database connection, the second a resource which holds the result of a query.

The cURL extension defines another two resource types:

… a cURL handle and a cURL multi handle.

When var_dumped, the values look like this:

$resource = curl_init();
var_dump($resource);

resource(1) of type (curl)

That’s all most resources are, a numeric identifier ((1)) of a certain type ((curl)).

You carry these resources around and pass them to different functions for which such a resource means something. Typically these functions allocate certain data in the background and a resource is just a reference which they use to keep track of this data internally.

The «… expects parameter 1 to be resource, boolean given» error is typically the result of an unchecked operation that was supposed to create a resource, but returned false instead. For instance, the fopen function has this description:

Return Values

Returns a file pointer resource on success, or FALSE on error.

So in this code, $fp will either be a resource(x) of type (stream) or false:

$fp = fopen(...);

If you do not check whether the fopen operation succeed or failed and hence whether $fp is a valid resource or false and pass $fp to another function which expects a resource, you may get the above error:

$fp   = fopen(...);
$data = fread($fp, 1024);

Warning: fread() expects parameter 1 to be resource, boolean given

You always need to error check the return value of functions which are trying to allocate a resource and may fail:

$fp = fopen(...);

if (!$fp) {
    trigger_error('Failed to allocate resource');
    exit;
}

$data = fread($fp, 1024);

Related Errors:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given

Warning: Illegal string offset ‘XXX’

This happens when you try to access an array element with the square bracket syntax, but you’re doing this on a string, and not on an array, so the operation clearly doesn’t make sense.

Example:

$var = "test";
echo $var["a_key"];

If you think the variable should be an array, see where it comes from and fix the problem there.

Code doesn’t run/what looks like parts of my PHP code are output

If you see no result from your PHP code whatsoever and/or you are seeing parts of your literal PHP source code output in the webpage, you can be pretty sure that your PHP isn’t actually getting executed. If you use View Source in your browser, you’re probably seeing the whole PHP source code file as is. Since PHP code is embedded in <?php ?> tags, the browser will try to interpret those as HTML tags and the result may look somewhat confused.

To actually run your PHP scripts, you need:

a web server which executes your script
to set the file extension to .php, otherwise the web server won’t interpret it as such*
to access your .php file via the web server

_{* Unless you reconfigure it, everything can be configured.}

This last one is particularly important. Just double clicking the file will likely open it in your browser using an address such as:

file://C:/path/to/my/file.php

This is completely bypassing any web server you may have running and the file is not getting interpreted. You need to visit the URL of the file on your web server, likely something like:

http://localhost/my/file.php

You may also want to check whether you’re using short open tags <? instead of <?php and your PHP configuration has turned short open tags off.

Also see PHP code is not being executed, instead code shows on the page

Warning: Array to string conversion

Notice: Array to string conversion

(A notice until PHP 7.4, since PHP 8.0 a warning)

This simply happens if you try to treat an array as a string:

$arr = array('foo', 'bar');

echo $arr;  // Notice: Array to string conversion
$str = 'Something, ' . $arr;  // Notice: Array to string conversion

An array cannot simply be echo‘d or concatenated with a string, because the result is not well defined. PHP will use the string «Array» in place of the array, and trigger the notice to point out that that’s probably not what was intended and that you should be checking your code here. You probably want something like this instead:

echo $arr[0];  // displays foo
$str = 'Something ' . join(', ', $arr); //displays Something, foo, bar

Or loop the array:

foreach($arr as $key => $value) {
    echo "array $key = $value";
    // displays first: array 0 = foo
    // displays next:  array 1 = bar
}

If this notice appears somewhere you don’t expect, it means a variable which you thought is a string is actually an array. That means you have a bug in your code which makes this variable an array instead of the string you expect.

Warning: mysql_connect(): Access denied for user ‘name’@’host’

This warning shows up when you connect to a MySQL/MariaDB server with invalid or missing credentials (username/password). So this is typically not a code problem, but a server configuration issue.

See the manual page on mysql_connect("localhost", "user", "pw") for examples.
Check that you actually used a $username and $password.
- It’s uncommon that you gain access using no password — which is what happened when the Warning: said (using password: NO).
- Only the local test server usually allows to connect with username root, no password, and the test database name.
- You can test if they’re really correct using the command line client:
  mysql --user="username" --password="password" testdb
- Username and password are case-sensitive and whitespace is not ignored. If your password contains meta characters like $, escape them, or put the password in single quotes.
- Most shared hosting providers predeclare mysql accounts in relation to the unix user account (sometimes just prefixes or extra numeric suffixes). See the docs for a pattern or documentation, and CPanel or whatever interface for setting a password.
- See the MySQL manual on Adding user accounts using the command line. When connected as admin user you can issue a query like:
  CREATE USER 'username'@'localhost' IDENTIFIED BY 'newpassword';
- Or use Adminer or WorkBench or any other graphical tool to create, check or correct account details.
- If you can’t fix your credentials, then asking the internet to «please help» will have no effect. Only you and your hosting provider have permissions and sufficient access to diagnose and fix things.
Verify that you could reach the database server, using the host name given by your provider:
ping dbserver.hoster.example.net
- Check this from a SSH console directly on your webserver. Testing from your local development client to your shared hosting server is rarely meaningful.
- Often you just want the server name to be "localhost", which normally utilizes a local named socket when available. Othertimes you can try "127.0.0.1" as fallback.
- Should your MySQL/MariaDB server listen on a different port, then use "servername:3306".
- If that fails, then there’s a perhaps a firewall issue. (Off-topic, not a programming question. No remote guess-helping possible.)
When using constants like e.g. DB_USER or DB_PASSWORD, check that they’re actually defined.
- If you get a "Warning: Access defined for 'DB_USER'@'host'" and a "Notice: use of undefined constant 'DB_PASS'", then that’s your problem.
- Verify that your e.g. xy/db-config.php was actually included and whatelse.
Check for correctly set GRANT permissions.
- It’s not sufficient to have a username+password pair.
- Each MySQL/MariaDB account can have an attached set of permissions.
- Those can restrict which databases you are allowed to connect to, from which client/server the connection may originate from, and which queries are permitted.
- The «Access denied» warning thus may as well show up for mysql_query calls, if you don’t have permissions to SELECT from a specific table, or INSERT/UPDATE, and more commonly DELETE anything.
- You can adapt account permissions when connected per command line client using the admin account with a query like:
  GRANT ALL ON yourdb.* TO 'username'@'localhost';
If the warning shows up first with Warning: mysql_query(): Access denied for user ''@'localhost' then you may have a php.ini-preconfigured account/password pair.
- Check that mysql.default_user= and mysql.default_password= have meaningful values.
- Oftentimes this is a provider-configuration. So contact their support for mismatches.
Find the documentation of your shared hosting provider:
- e.g. HostGator, GoDaddy, 1and1, DigitalOcean, BlueHost, DreamHost, MediaTemple, ixWebhosting, lunarhosting, or just google yours´.
- Else consult your webhosting provider through their support channels.
Note that you may also have depleted the available connection pool. You’ll get access denied warnings for too many concurrent connections. (You have to investigate the setup. That’s an off-topic server configuration issue, not a programming question.)
Your libmysql client version may not be compatible with the database server. Normally MySQL and MariaDB servers can be reached with PHPs compiled in driver. If you have a custom setup, or an outdated PHP version, and a much newer database server, or significantly outdated one — then the version mismatch may prevent connections. (No, you have to investigate yourself. Nobody can guess your setup).

More references:

Serverfault: mysql access denied for ‘root’@’name of the computer’
Warning: mysql_connect(): Access denied
Warning: mysql_select_db() Access denied for user »@’localhost’ (using password: NO)
Access denied for user ‘root’@’localhost’ with PHPMyAdmin

Btw, you probably don’t want to use mysql_* functions anymore. Newcomers often migrate to mysqli, which however is just as tedious. Instead read up on PDO and prepared statements.
$db = new PDO("mysql:host=localhost;dbname=testdb", "username", "password");

Deprecated: Array and string offset access syntax with curly braces is deprecated

String offsets and array elements could be accessed by curly braces {} prior to PHP 7.4.0:

$string = 'abc';
echo $string{0};  // a

$array = [1, 2, 3];
echo $array{0};  // 1

This has been deprecated since PHP 7.4.0 and generates a warning:

Deprecated: Array and string offset access syntax with curly braces is deprecated

You must use square brackets [] to access string offsets and array elements:

$string = 'abc';
echo $string[0];  // a

$array = [1, 2, 3];
echo $array[0];  // 1

The RFC for this change links to a PHP script which attempts to fix this mechanically.

Warning: Division by zero

The warning message ‘Division by zero’ is one of the most commonly asked questions among new PHP developers. This error will not cause an exception, therefore, some developers will occasionally suppress the warning by adding the error suppression operator @ before the expression. For example:

$value = @(2 / 0);

But, like with any warning, the best approach would be to track down the cause of the warning and resolve it. The cause of the warning is going to come from any instance where you attempt to divide by 0, a variable equal to 0, or a variable which has not been assigned (because NULL == 0) because the result will be ‘undefined’.

To correct this warning, you should rewrite your expression to check that the value is not 0, if it is, do something else. If the value is zero you either should not divide, or you should change the value to 1 and then divide so the division results in the equivalent of having divided only by the additional variable.

if ( $var1 == 0 ) { // check if var1 equals zero
    $var1 = 1; // var1 equaled zero so change var1 to equal one instead
    $var3 = ($var2 / $var1); // divide var1/var2 ie. 1/1
} else {
    $var3 = ($var2 / $var1); // if var1 does not equal zero, divide
}

Related Questions:

warning: division by zero
Warning: Division By Zero Working on PHP and MySQL
Division by zero error in WordPress Theme
How to suppress the “Division by zero” error
How to catch a division by zero?

Strict Standards: Non-static method [<class>::<method>] should not be called statically

Occurs when you try to call a non-static method on a class as it was static, and you also have the E_STRICT flag in your error_reporting() settings.

Example :

class HTML {
   public function br() {
      echo '<br>';
   }
}

HTML::br() or $html::br()

You can actually avoid this error by not adding E_STRICT to error_reporting(), eg

error_reporting(E_ALL & ~E_STRICT);

since as for PHP 5.4.0 and above, E_STRICT is included in E_ALL [ref]. But that is not adviceable. The solution is to define your intended static function as actual static :

public static function br() {
  echo '<br>';
}

or call the function conventionally :

$html = new HTML();
$html->br();

php — Parse error: syntax error, unexpected string content «», expecting «-» or identifier or variable or number

Answer

Solution:

There is indeed a problem with the syntax of the 146th line, but that isn’t the primary problem you have here.

You should read up on this question for a better way to query mysql from inside PHP, even after fixing the syntax problem, you’re open to SQL injection, allowing any visitor to delete your entire database: How can I prevent SQL injection in PHP?

Changing the query into a prepared statement solves your syntax problem along the way.

I’m hesitant to include the quickfix for the syntax, as i DO NOT think you should use this fix and continue onwards (i really think you should follow the advice in the link above), but it is:

Additional Information:

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About the technologies asked in this question

PHP (from the English Hypertext Preprocessor — hypertext preprocessor) is a scripting programming language for developing web applications. Supported by most hosting providers, it is one of the most popular tools for creating dynamic websites. The PHP scripting language has gained wide popularity due to its processing speed, simplicity, cross-platform, functionality and distribution of source codes under its own license.
https://www.php.net/

MySQL

DBMS is a database management system. It is designed to change, search, add and delete information in the database. There are many DBMSs designed for similar purposes with different features. One of the most popular is MySQL. It is a software tool designed to work with relational SQL databases. It is easy to learn even for site owners who are not professional programmers or administrators. MySQL DBMS also allows you to export and import data, which is convenient when moving large amounts of information.
https://www.mysql.com/

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Источник

Отладчик указывает на синтаксическую ошибку, а ее там нет почему?

Parse error: syntax error, unexpected » (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in C:OpenServerdomainslocalhostsignup.php on line 48

Сам код:

<?php
	require_once 'header.php';
	
	echo <<<_END
	<script>
	function checkUser(user)
	{
		if (user.value == '')
		{
			O('info').innerHTML = ''
			return
		}
		
		params = "user=" + user.value
		request = new ajaxRequest()
		request.open("POST", "checkuser.php", true)
		request.setRequestHeader("Content-type","application/x-www-form-urlencoded")
		request.setRequestHeader("Connection", "close")
		
		request.onreadystatechange = function()
		{
				if (this.readyState == 4)
					if (this.status == 200)
						if (this.responseText != null)
							O('info').innerHTML = this.responseText
		}
		request.send(params)
	}
	
	function ajaxRequest()
	{
		try {var request = new XMLHttpRequest()}
		catch(e1){
			try {request = new ActiveXObject("Msxml2.XMLHTTP")}
			catch(e2)
				try {request = new ActiveXObject("Microsoft.XMLHTTP")}
				catch(e3){
					request = false
		}	}	}
		return request
	}
	</script>
	<div class='main'><h3>Please enter your details to sign up</h3>
	// Введите, пожайлуста, свои регистрационные данные
	_END;
	
	$error = $user = $pass = "";
	if (isset($_POST['user'])) destroySession();
	
	if (isset($_POST['user']))
	{
		$user = sanitizeString($_POST['user']);
		$pass = sanitizeString($_POST['pass']);
		
		if ($user == "" || $pass == "")
			$error = "Данные введены не во все поля<br><br>";
		else
		{
			$result = queryMysql("SELECT * FROM members WHERE user='$user'");
			if ($result->num_rows)
				$error = "Такое имя уже существует<br><br>";
			else
			{
				queryMysql("INSERT INTO members VALUES('$user','$pass')");
				die("<h4>Account created</h4>Please Log in.<br><br>");
				// Учетная запись создана,
				// пожайлуста, войдите на сайт
			}
		}
	}
	
	echo <<<_END
	<form method=;post' action='signup.php'>$error
		<span class='fieldname'>Username</span>
		<input type='text' maxlength='16' name='user' value='$user' onBlur='checkUser(this)'><span id='info'></span><br>
		<span class='fieldname'>Password</span>
		<input type='text' maxlength='16' name='pass' value='$pass'><br>
		_END;
		?>
		
		<span class='fieldname'>&nbsp;</span>
		<input type='submit' value='Sign up' />
		</form></div><br>
		</body>
	</html>
?>

Источник

svva643

6 / 4 / 2

Регистрация: 08.05.2021

Сообщений: 133

31.07.2021, 12:39. Показов 2251. Ответов 5

Метки нет (Все метки)

Ошибка Parse error: syntax error, unexpected » (T_ENCAPSED_AND_WHITESPACE), expecting ‘-‘ or identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in I:OpenServerdomainskeeppixel.tkdevinfoup.php on line 6. вылезает когда я пытаюсь создать папку или переменную с заданным человеком именем в строке допустим есть пост запрос он приравнен data

PHP

1	$data = $_POST

получатся переменная

PHP

1	$folder = "I:openserverdomainskeeppixel.tk/chgm/$pe/$data['namegame']/'".$data['namegame']."'.txt";

Далее mk и так далее, но ошибка в $data[‘namegame’] это не string и я пробовал по разному. дин привер как я пробовал выше.
Отрывок всего php кода:

PHP

<?php
require "db.php";
  $data = $_POST;
  $pe = $_SESSION['logged_user']->id;
  $per = R::load('users', $pe);
  $foldeaar = "I:openserverdomainskeeppixel.tk/chgm/$pe/$data['namegame']/'".$data['namegame']."'.txt";
  $rek = $_COOKIE["IDC"];
 if( isset($_SESSION['logged_user']))
 {
     
 }
 
 
if(isset($data['do_signup']))
{
  
  
    if( empty($errors) )      
      {
              
       $user = R::findOne('gamecheck', 'hash = ?', array($rek));
       if( $user )
       {
            header('Location: http://keeppixel.tk/devgameup.php');
          $users = R::load('gamecheck', $user->id);
          // Обращаемся к свойству объекта и назначаем ему новое значение
          $users->opic = $data['opicgame'];
          $users->game = $data['namegame'];
          $users->janr = $data['janr'];
          // Сохраняем объект
          R::store($users);
           
       }else 
       {
           $errors[] = 'Ошибка, неправильный хэш. Рассинхронизация, пожалусьа вернитесь назад и попробу1те снова';
       }
   
      }
    if (!file_exists($foldeaar)) {
  $folderi = "I:openserverdomainskeeppixel.tk/chgm/$pe/$data['namegame']"; // путь и имя новой папки
  mkdir($folderi, 0777, true); 
  $afolder = "I:openserverdomainskeeppixel.tk/chgm/$pe/$data['namegame']/video"; // путь и имя новой папки
  mkdir($afolder, 0777, true); 
  $foldera = "I:openserverdomainskeeppixel.tk/chgm/$pe/$data['namegame']/img"; // путь и имя новой папки
  mkdir($foldera, 0777, true); 
  $text = "$data['namegame'] не трогать это пакет ресурсов";
 
// открываем файл, если файл не существует,
//делается попытка создать его
$fp = fopen("I:openserverdomainskeeppixel.tk/chgm/$pe/$data['namegame']/$data['namegame'].txt", "w");
 
// записываем в файл текст
fwrite($fp, $text);
 
// закрываем
fclose($fp);
  }
       $errors = array();
       if( trim($data['namegame']) == '' )
      {
          $errors[] = 'Введите название игры';
      }
      if( trim($data['opicgame']) == '' )
      {
          $errors[] = 'Введите описание для игры';
      }   
      if( trim($data['janr']) == '' )
      {
          $errors[] = 'Введите жанр игру';
      }
  $vu = "$pe+$file_name";
 
  
} 
 
?>

Заранее спасибо за помощь !!

__________________
Помощь в написании контрольных, курсовых и дипломных работ, диссертаций здесь

Programming

Эксперт

94731 / 64177 / 26122

Регистрация: 12.04.2006

Сообщений: 116,782

31.07.2021, 12:39

Ответы с готовыми решениями:

Parse error: syntax error, unexpected » (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING
Помогите пжлст решить проблему с этой ошибкой!

<?php

$mytext = "<?php …

Ошибка «Parse error: syntax error, unexpected ‘»‘, expecting identifier (T_STRING) or variable (T_VARIABLE)
Пытаюсь сделать фильтр с помощью eval()
foreach ($podbors as $podbor){

if (isset($podbor)) {
…

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING
Здравствуйте, при попытке объединить регистрацию на DLE и IPB у меня вылезает ошибка:

Parse…

Ошибка Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING
Ясно, что проблема с кавычками, но как исправить :о
Parse error: syntax error, unexpected…

~~Camy~~

Заблокирован

31.07.2021, 12:47

Сообщение от svva643

openserverdomainskeeppixel.tk/chgm/$pe/$data[‘namegame’]/’

Слэши в разные стороны) Переходи на Линукс)

6 / 4 / 2

Регистрация: 08.05.2021

Сообщений: 133

31.07.2021, 13:15

[ТС]

я строю сервер но тестовый режим на винде, типа в этом ошибка

ft4l

Невнимательный

2329 / 684 / 251

Регистрация: 08.02.2013

Сообщений: 4,770

Записей в блоге: 2

31.07.2021, 15:12

Сообщение было отмечено svva643 как решение

Решение

Сообщение от svva643

типа в этом ошибка

Типа в слешах … непохоже

Сообщение от svva643

fopen(«I:openserverdomainskeeppixel.tk/chgm/$pe/$data[‘namegame’]/$data[‘namegame’].txt», «w»);

o d k не интерпретируются никак

Двойные кавычки
Если строка заключена в двойные кавычки («), PHP распознает следующие управляющие последовательности специальных символов:

Управляющие последовательности

Последовательность	Значение
n	новая строка (LF или 0x0A (10) в ASCII)
r	возврат каретки (CR или 0x0D (13) в ASCII)
t	горизонтальная табуляция (HT или 0x09 (9) в ASCII)
v	вертикальная табуляция (VT или 0x0B (11) в ASCII)
e	escape-знак (ESC или 0x1B (27) в ASCII)
f	подача страницы (FF или 0x0C (12) в ASCII)
\	обратная косая черта
$	знак доллара
»	двойная кавычка
[0-7]{1,3}	последовательность символов, соответствующая регулярному выражению символа в восьмеричной системе счисления, который молча переполняется, чтобы поместиться в байт (т.е. «400» === «00»)
x[0-9A-Fa-f]{1,2}	последовательность символов, соответствующая регулярному выражению символа в шестнадцатеричной системе счисления
u{[0-9A-Fa-f]+}	последовательность символов, соответствующая регулярному выражению символа Unicode, которая отображается в строка в представлении UTF-8

Но с апострофами намного проще строки… там только ‘ и \
… остальное будет как есть

Сообщение от svva643

Parse error

PHP

$ cat <<x | /c/srv/usr/php74/php
<?php
$arr = ['key'=>'value'];
 
echo '  ', var_export("...$arr[key]...");
echo '  ', var_export("...{$arr['key']}...");
 
// echo '  ', var_export("...$arr['key']...");// Parse error
 
x
  '...value...'  '...value...'

6 / 4 / 2

Регистрация: 08.05.2021

Сообщений: 133

31.07.2021, 15:29

[ТС]

Спасибо я че то затупил я пробовал так только в одном месте на строке а видишь перед txt папка с таким же название вот я его скинул поэтому ошибка оставалась на этой строчке

ft4l

31.07.2021, 15:30

Parse error: syntax error, unexpected » (T_ENCAPSED_AND_WHITESPACE), expecting ‘-‘ or identifier (T_STRING) or variable

Не по теме:

Сообщение от x_lab

x[0-9A-Fa-f]{1,2}

Одно время в notepad++ не смог найти способ в «заготовки» QuickText добавлять nn
и добавлял «xAxA» . PHP_EOL . PHP_EOL как-то неоднозначно ,)

И как-то однажды «знакомая» увидела что я долго ковыряюсь в каком-то тексте , где в каждой строке xAxA , xAxA
Посмотрела как на психа ,) интересуясь сам-ли я это сочиняю

IT_Exp

Эксперт

87844 / 49110 / 22898

Регистрация: 17.06.2006

Сообщений: 92,604

31.07.2021, 15:30

Ошибка в php — Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in CODE on line 9
Доброго времени суток! Есть код, в нем где-то есть ошибка (вроде кавычки), но знаний php у меня 0,…

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in Z:homescript.uawwwscript_1.ph
имеется код
$res = mysql_query("SELECT * FROM news");
$count = mysql_num_rows($res);
$res_2 =…

syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in
ошибка Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or…

Syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING
Установил денвер. Запускаю файл

<html>
<head>
<meta charset="utf-8">
</head>
<body> …

Ошибка Parse error: syntax error, unexpected T_STRING, expecting T_CONSTANT_ENCAPSED_STRIN
При запуске локального сервера, выдает эту ошибку.
Parse error: syntax error, unexpected…

Parse error: syntax error, unexpected ‘Бизнес’ (T_STRING), expecting ‘,’ or ‘)’
Добрый день, есть данная ошибка Parse error: syntax error, unexpected ‘Бизнес’ (T_STRING),…

Искать еще темы с ответами

Или воспользуйтесь поиском по форуму:

Источник

Describe the bug
When pushing data form fluent-plugin-grafana-loki the API call results in error:
parse error at line 1, col 2: syntax error: unexpected }, expecting IDENTIFIER

This also occurs when trying to replicate the API call via curl.

To Reproduce
Steps to reproduce the behavior:

Started Loki (FROM grafana/loki:master-eb7b495)
Started fluentd with loki plugin fluent-plugin-grafana-loki-1.2.3
Push some logs to loki (via fluentd or curl)

# test.json
{
  "streams": [
    {
      "labels": {
        "a": 123
      },
      "entries": [
        {
          "ts": "2019-12-02T11:16:12.000000Z",
          "line": "{"OriginStatus":100}"
        }
      ]
    }
  ]
}

# api call
curl -XPOST -u loki:PassWord  -H "Content-Type: application/json" --data-binary "@./test.json"  https://loki.example.com/loki/api/v1/push
parse error at line 1, col 2: syntax error: unexpected }, expecting IDENTIFIER

I also tried messing around with the test.json, giving different line and labels (escaping) but still the same error.

This is a shorter json, the actual json, which is causing the error originates from the fluentd pluign.

Expected behavior
I expected the data to be stored in loki -> successful API call

Environment:

Infrastructure: docker, debian, baremetal
Deployment tool: rancher, docker-compose

Loki logs just contain something like this:

today at 1:06 PM  level=info ts=2019-12-02T12:06:46.528852361Z caller=table_manager.go:363 msg="creating table" table=index_2555
today at 1:06 PM  level=info ts=2019-12-02T12:06:46.528931231Z caller=table_manager.go:363 msg="creating table" table=index_2559
today at 1:06 PM  level=info ts=2019-12-02T12:06:46.529055099Z caller=table_manager.go:363 msg="creating table" table=index_2585
today at 1:06 PM  level=info ts=2019-12-02T12:06:46.529134623Z caller=table_manager.go:363 msg="creating table" table=index_2589
today at 1:06 PM  level=info ts=2019-12-02T12:06:46.529261024Z caller=table_manager.go:363 msg="creating table" table=index_2523
today at 1:06 PM  level=info ts=2019-12-02T12:06:46.529339896Z caller=table_manager.go:363 msg="creating table" table=index_2530
today at 1:06 PM  level=info ts=2019-12-02T12:06:46.529557986Z caller=table_manager.go:409 msg="provisioned throughput on table, skipping" table=index_2604 read=0 write=0

The actual log error:

today at 1:07 PM  2019-12-02 12:07:50 +0000 [warn]: #0 {"streams":[{"labels":"{}","entries":[{"ts":"2019-12-02T12:07:39.000000Z","line":"{"BackendName":"backend-loki-fluetd","ClientHost":"10.42.53.79","Duration":1822,"FrontendName":"Host-loki-example-net","OriginContentSize":79,"OriginDuration":1738717,"OriginStatus":400,"Overhead":83599,"RequestAddr":"loki.example.net","RequestContentSize":3495,"RequestCount":2453697,"RequestHost":"loki.example.net","RequestLine":"POST /loki/api/v1/push HTTP/1.1","RetryAttempts":0,"StartLocal":"2019-12-02T12:07:39.064023992Z","downstream_Content-Length":"79","downstream_Content-Type":"text/plain; charset=utf-8","downstream_Date":"Mon, 02 Dec 2019 12:07:39 GMT","downstream_X-Content-Type-Options":"nosniff","level":"info","msg":"","origin_Content-Length":"79","origin_Content-Type":"text/plain; charset=utf-8","origin_Date":"Mon, 02 Dec 2019 12:07:39 GMT","origin_X-Content-Type-Options":"nosniff","request_Accept":"*/*","request_Accept-Encoding":"gzip;q=1.0,deflate;q=0.6,identity;q=0.3","request_Authorization":"Basic XXXXXXXXXXXXXXXXXX","request_Content-Length":"3495","request_Content-Type":"application/json","request_User-Agent":"Ruby"}"},{"ts":"2019-12-02T12:07:39.000000Z","line":"{"BackendName":"backend-app-api-app-app-api-app-servicewithauth","ClientHost":"82.119.96.82","Duration":2697,"FrontendName":"app-api-app-app-api-app-servicewithauth","OriginContentSize":40,"OriginDuration":2586404,"OriginStatus":400,"Overhead":111158,"RequestAddr":"app.example.net","RequestContentSize":0,"RequestCount":2453698,"RequestHost":"app.example.net","RequestLine":"GET /socket.io/?__sails_io_sdk_version=0.14.0&EIO=3&transport=polling HTTP/2.0","RetryAttempts":0,"StartLocal":"2019-12-02T12:07:39.47041769Z","downstream_Access-Control-Allow-Origin":"*","downstream_Content-Type":"application/json","downstream_Date":"Mon, 02 Dec 2019 12:07:39 GMT","level":"info","msg":"","origin_Access-Control-Allow-Origin":"*","origin_Content-Type":"application/json","origin_Date":"Mon, 02 Dec 2019 12:07:39 GMT","request_Accept":"*/*","request_Accept-Encoding":"gzip","request_Authorization":"Basic XXXXXXXXXXXXXXXXXX","request_User-Agent":"okhttp/4.2.2"}"},{"ts":"2019-12-02T12:07:45.000000Z","line":"{"BackendName":"backend-app-api-app-app-api-app-servicewithauth","ClientHost":"82.119.96.82","Duration":2375,"FrontendName":"app-api-app-app-api-app-servicewithauth","OriginContentSize":40,"OriginDuration":2272545,"OriginStatus":400,"Overhead":102813,"RequestAddr":"app.example.net","RequestContentSize":0,"RequestCount":2453699,"RequestHost":"app.example.net","RequestLine":"GET /socket.io/?__sails_io_sdk_version=0.14.0&EIO=3&transport=polling HTTP/2.0","RetryAttempts":0,"StartLocal":"2019-12-02T12:07:45.234770579Z","downstream_Access-Control-Allow-Origin":"*","downstream_Content-Type":"application/json","downstream_Date":"Mon, 02 Dec 2019 12:07:45 GMT","level":"info","msg":"","origin_Access-Control-Allow-Origin":"*","origin_Content-Type":"application/json","origin_Date":"Mon, 02 Dec 2019 12:07:45 GMT","request_Accept":"*/*","request_Accept-Encoding":"gzip","request_Authorization":"Basic XXXXXXXXXXXXXXXXXX","request_User-Agent":"okhttp/4.2.2"}"}]}]}
today at 1:08 PM  2019-12-02 12:08:01 +0000 [warn]: #0 failed to POST https://loki.example.net/loki/api/v1/push (400 Bad Request parse error at line 1, col 2: syntax error: unexpected }, expecting IDENTIFIER

Источник

Warning: Cannot modify header information — headers already sent

Fatal error: Call to a member function … on a non-object

«Notice: Undefined Index», or «Warning: Undefined array key»

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given

Fatal error: Using $this when not in object context

Fatal error: Call to undefined function XXX

Fatal error: Can’t use function return value in write context

MySQL: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near … at line …

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE

The error case:

Possible fixes:

Fatal error: Allowed memory size of XXX bytes exhausted (tried to allocate XXX bytes)

Warning: [function]: failed to open stream: [reason]

Notice: Undefined variable

Notice: Undefined property

Notice: Use of undefined constant XXX — assumed ‘XXX’

Warning: Use of undefined constant XXX — assumed ‘XXX’ (this will throw an Error in a future version of PHP)

Error: Undefined constant XXX

Fatal error: Cannot redeclare class [class name]

Fatal error: Cannot redeclare [function name]

Parse error: syntax error, unexpected T_VARIABLE

Notice: Uninitialized string offset: *

Notice: Trying to get property of non-object error

Parse error: syntax error, unexpected ‘[‘

Variation 1

Variation 2

Warning: [function] expects parameter 1 to be resource, boolean given

Warning: Illegal string offset ‘XXX’

Code doesn’t run/what looks like parts of my PHP code are output

Warning: Array to string conversion

Notice: Array to string conversion

Warning: mysql_connect(): Access denied for user ‘name’@’host’

Deprecated: Array and string offset access syntax with curly braces is deprecated

Warning: Division by zero

Strict Standards: Non-static method [<class>::<method>] should not be called statically

php — Parse error: syntax error, unexpected string content «», expecting «-» or identifier or variable or number

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Notice: Uninitialized string offset: `*`