Fatal error continue not in the loop or switch context in

<?phpfor( $i = 0; $i < 3; ++ $i )
    {
        echo ' [', $i, '] ';
        switch( $i )
        {
            case 0: echo 'zero'; break;
            case 1: echo 'one' ; XXXX;
            case 2: echo 'two' ; break;
        }
        echo ' <' , $i, '> ';
    }?>

For XXXX I filled in

- continue 1
- continue 2
- break 1
- break 2

and observed the different results.  This made me come up with the following one-liner that describes the difference between break and continue:

continue resumes execution just before the closing curly bracket ( } ), and break resumes execution just after the closing curly bracket.

Corollary: since a switch is not (really) a looping structure, resuming execution just before a switch's closing curly bracket has the same effect as using a break statement.  In the case of (for, while, do-while) loops, resuming execution just prior their closing curly brackets means that a new iteration is started --which is of course very unlike the behavior of a break statement.

In the one-liner above I ignored the existence of parameters to break/continue, but the one-liner is also valid when parameters are supplied.

<?php

$stack = array('first', 'second', 'third', 'fourth', 'fifth');

foreach(

$stack AS $v){

    if($v == 'second')continue;

    if($v == 'fourth')break;

    echo $v.'<br>';

}

/*

first

third

*/

$stack2 = array('one'=>'first', 'two'=>'second', 'three'=>'third', 'four'=>'fourth', 'five'=>'fifth');

foreach($stack2 AS $k=>$v){

    if($v == 'second')continue;

    if($k == 'three')continue;

    if($v == 'fifth')break;

    echo $k.' ::: '.$v.'<br>';

}

/*

one ::: first

four ::: fourth

*/

?>

<?php
$arr = array(1, 2, 3);
foreach($arr as $number) {
  if($number == 2) {
    continue;
  }
  print $number;
}
?>

<?
    for ($i = 0;$i<3;$i++) {
        echo "Start Of I loopn";
        for ($j=0;;$j++) {

                        if ($j >= 2) continue 2; // This "continue" applies to the "$i" loop
            echo "I : $i J : $j"."n";
        }
        echo "Endn";
    }
?>

The output here is:
Start Of I loop
I : 0 J : 0
I : 0 J : 1
Start Of I loop
I : 1 J : 0
I : 1 J : 1
Start Of I loop
I : 2 J : 0
I : 2 J : 1

For more information, see the php manual's entry for the 'break' statement.

//page1.php
for($x=0;$x<10;$x++)
   {
    include('page2.php');   
}

//page2.php

if($x==5)
    continue;
else
   print $x;

it should print

"012346789" no five, but it produces an error:

Cannot break/continue 1 level in etc.

So the only way to give the control back to the loop-operation  in page1.php would be a return.

<?php
$i = 0;
while ($i++ < 5) {
    while ($i % 2 === 0) {
        echo "$i is even. n";
        ####
    }
    echo "$i is odd. n";
}
?>

Where ####:
- break: outputs both '$i is even' and '$i is odd' for even numbers.
- continue: infinite loop as soon as it evaluates true for the first even number.
- break 2: as soon as it runs, it exits from both loops.
- continue 2: outputs numbers correctly.

What I understand from this is that break will exit current looping structure and will keep running outer loop code. Continue will make loop get back to evaluation, and will iterate over itself until it evaluates to false. Break 2 will exit 2 levels, which in this case will stop the iteration altogether. Continue 2 will evaluate not the current loop (level 1 so to speak), but the outer loop in this case.

<?php
// Outputs "1 ".
$i = 0;
while ($i == 0) {
    $i++;
    echo "$i ";
    if ($i == 1) continue;
}// Outputs "1 2 ".
$i = 0;
do {
    $i++;
    echo "$i ";
    if ($i == 2) continue;
} while ($i == 1);
?>

Both code snippets would behave exactly the same without continue.

This example skips the value of 4:

<?php
for ($x = 0; $x < 10; $x++) {
  if ($x == 4) {
    continue;
  }
  echo "The number is: $x <br>";
}
?>

11,13,17,19,23,29,

PHP Fatal error:  'continue' operator accepts only positive numbers

(same is true for break).

It is said in manually:
continue also accepts an optional numeric argument which tells it how many levels of enclosing loops it should .

In order to understand better this,An example for that:
<?php/*continue also accepts an optional numeric argument which
    tells it how many levels of enclosing loops it should skip.*/for($k=0;$k<2;$k++)
{//First loopfor($j=0;$j<2;$j++)
    {//Second loopfor($i=0;$i<4;$i++)
      {//Third loop
    if($i>2)
    continue 2;// If $i >2 ,Then it skips to the Second loop(level 2),And starts the next step,
    echo "$in";
       }

    }

}

?>

Merry's christmas :)

    With regards,Hossein

I've got the same problem as Greg
and now it works very fine by using
return() instead of continue.

It seems, that you have to use return()
if you have a file included and
you want to continue with the next loop

If you need to correct such error cases as part of an upgrade, you may need to substitute either an exit or return to maintain the existing behavior of such legacy code.

<?phpclass ok {

    function

foo() {
        echo "startn";

        for (

$i = 0; $i < 5; $i++) {
            echo "beforen";
            $this->bar($i);
            echo "aftern";
        }

        echo

"finishn";
    }

    function

bar($i) {
        echo "inside iteration $in";

                if (

$i == 3) {
            echo "continuingn";
            continue;
        }

        echo

"inside after $in";
    }
}$ex = new ok();$ex->foo();?>

sh> php56 continue.php
start
before
inside iteration 0
inside after 0
after
before
inside iteration 1
inside after 1
after
before
inside iteration 2
inside after 2
after
before
inside iteration 3
continuing
PHP Fatal error:  Cannot break/continue 1 level in continue.php on line 22
PHP Stack trace:
PHP   1. {main}() continue.php:0
PHP   2. ok->foo() continue.php:31
PHP   3. ok->bar() continue.php:10

sh> php70 continue.php
PHP Fatal error:  'continue' not in the 'loop' or 'switch' context in continue.php on line 22

Fatal error: 'continue' not in the 'loop' or 'switch' context in continue.php on line 22

    for (

$i = 0; $i < 5; $i++ ) {

        switch (

$i)
        {

            case

0:
                echo $i . "b";
                continue;
                echo $i . "a";
            case 1:   
                echo $i . "b";
                continue 2;
                echo $i . "a";
            case 2:   
                echo $i . "b";
                break;
                echo $i . "a";
            case 3:
                echo $i . "b";
                break 2;
                echo $i . "a";
            case 4:
                echo $i;

                    }

        echo

9;

    }

echo

"n";
echo"n";?>

This results in: 0b91b2b93b

It goes to show that in a switch statement break and continue are the same. But in loops break stops the loop completely and continue just stops executing the current iterations code and moves onto the next loop iteration.

"continue is used within looping structures to skip the rest of the current loop iteration"

Current functionality treats switch structures as looping in regards to continue.  It has the same effect as break.

The following code is an example:

<?php
for ($i1 = 0; $i1 < 2; $i1++) {
  // Loop 1.
  for ($i2 = 0; $i2 < 2; $i2++) {
    // Loop 2.
    switch ($i2 % 2) {
      case 0:
        continue;
        break;
    }
    print '[' . $i2 . ']<br>';
  }
  print $i1 . '<br>';
}
?>

This outputs the following:
[0]
[1]
0
[0]
[1]
1

Switch is documented as a block of if...elseif... statements, so you might expect the following output:
[1]
0
[1]
1

This output requires you to either change the switch to an if or use the numerical argument and treat the switch as one loop.

the script will output "2" because the missing semikolon causes that the "print"-call is executed only if the "if" statement is true. It has nothing to to with "what" the "print"-call would return or not return, but the returning value can cause to skip to the end of higher level Loops if any call is used that will return a bigger number than 1.

<?php
continue print "$in";
?>

because of the optional argument, the script will not run into a "unexpected T_PRINT" error. It will not run into an error, too, if the call after continue does return anything but a number.

i suggest to change it from:
because the return value of the print() call is int(1), and it will look like the optional numeric argument mentioned above.

to
because the print() call will look like the optional numeric argument mentioned above.