Error call to undefined function tests feature factory

<?php

namespace TestsFeatureHttpControllersAuth;

use IlluminateFoundationTestingRefreshDatabase;
use IlluminateFoundationTestingWithFaker;
use TestsTestCase;
use AppUser;

class LoginControllerTest extends TestCase
{
    use RefreshDatabase;
    /** @test */
    public function login_authenticates_and_redirects_user()
    {
        $user = factory(User::class)->create();

        $response = $this->post(route('login'), [
            'email' => $user->email,
            'password' => 'password'
        ]);

        $response->assertRedirect(route('home'));
        $this->assertAuthenticatedAs($user);
    }
}

1) TestsFeatureHttpControllersAuthLoginControllerTest::login_authenticates_and_redirects_user
Error: Call to undefined function TestsFeatureHttpControllersAuthfactory()

1

684 votes

4 answers

In laravel I have written the following test:

    public function testUserCanCreateAssortment()
    {
        $this->signIn();

        $this->get('/assortments/create')->assertStatus(200);

        $this->followingRedirects()
            ->post('/assortments', $attributes = Assortment::factory()->raw())
            ->assertSee($attributes['title'])
            ->assertSee($attributes['description']);
    }
}

When I run it with the commandphpunit --filter testUserCanCreateItem I get the following error:

Error: Call to undefined function Testsfactory()

No idea what is causing it. I have looked at my factories and mytestcase.php but I could not find a solution. What am I doing wrong?

Mytestcase.php:

<?php

namespace Tests;

use IlluminateFoundationTestingTestCase as BaseTestCase;

abstract class TestCase extends BaseTestCase
{
    use CreatesApplication;

    protected function signIn($user = null)
    {
        $user = $user ?: User::factory()->create();

        $this->actingAs($user);

        return $user;
    }
}

Here the lines the error provides:

/var/www/tests/TestCase.php:13

/var/www/tests/Feature/ItemTest.php:29

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В laravel я написал следующий тест:

    public function testUserCanCreateAssortment()
    {
        $this->signIn();

        $this->get('/assortments/create')->assertStatus(200);

        $this->followingRedirects()
            ->post('/assortments', $attributes = Assortment::factory()->raw())
            ->assertSee($attributes['title'])
            ->assertSee($attributes['description']);
    }
}

Когда я запускаю его с помощью команды phpunit --filter testUserCanCreateItem, я получаю следующую ошибку:

Error: Call to undefined function Testsfactory()

Не знаю, что это вызывает. Я просмотрел свои заводы и свой testcase.php, но не смог найти решения. Что я делаю не так?

Мой testcase.php:

<?php

namespace Tests;

use IlluminateFoundationTestingTestCase as BaseTestCase;

abstract class TestCase extends BaseTestCase
{
    use CreatesApplication;

    protected function signIn($user = null)
    {
        $user = $user ?: User::factory()->create();

        $this->actingAs($user);

        return $user;
    }
}

Вот строки, содержащиеся в ошибке:

/var/www/tests/TestCase.php:13

/var/www/tests/Feature/ItemTest.php:29

Ребят нашел решение на свой ответ.

Мне нужно было добавить название модели в свой AssortmentFactory. Так я и сделал:

protected $model = Assortment::class;

И мне также нужно было импортировать модель, выполнив use App/Models/Assortment.

В laravel я написал следующий тест:…

Это решило мою проблему.

Возможно, вам придется указать, что вы используете другой класс.

Вы писали выше use Assortment или use Assortment?

Вы также можете использовать менеджер, чтобы найти класс, который вы используете

В Laravel 8 помощник factory больше не доступен. Класс модели testcase должен использовать свойство HasFactory, тогда вы можете использовать свою фабрику следующим образом:

testcase::factory()->count(50)->create();

Обратите внимание, что вам также следует обновить вызов фабрики пользователей: factory (‘App User’) -> create () -> id;

Вот соответствующая документация: https://laravel.com/docs/8.x/database- тестирование # create-models

Однако, если вы предпочитаете использовать фабрики стиля Laravel 7.x, вы можете использовать пакет laravel / legacy-factoryies. Вы можете установить его с помощью composer:

composer require laravel/legacy-factories

Many of you have encountered this error several times Fatal error: Call to undefined function function_name(). In today’s post, we are finding out how to unravel this error. But before we solve this problem, let’s understand how PHP evaluates the functions.

There are several ways to define functions and call them. Let’s say you write it in the function.php file and call it in the main.php file.

    // function.php
<?php
    namespace fooNamespace {
        function foo() {
            return "Calling foo"
        }
    }
?>
    // main.php
include function.php
<?php
    echo fooNamespacefoo();
?>

Namespaces are qualifiers that enable better management by grouping classes that work together to perform a task. It allows you to use the same name for multiple classes. It is important to know how PHP knows which element of the namespace is required by the code. PHP namespaces work quite a sort of a filesystem. There are 3 ways to access a file in a file system:

1. Relative file name such as fooBar.txt. It will resolve to fooDirectory/fooBar.txt where fooDirectory is the directory currently busy directory.
2. Relative path name such as subdirectory/fooBar.txt. It will resolve to fooDirectory/subdirectory/fooBar.txt.
3. Absolute path name such as /main/fooBar.txt. It will resolve to /main/fooBar.txt.

Namespaced elements in PHP follow an equivalent principle. For example, a class name can be specified in three ways:

1. Unqualified name/Unprefixed class name:Or,If the current namespace is foonamespace, it will always resolve to foonamespacefoo. If the code is a global, non-namespaced code, this resolves to foo.
2. Qualified name/Prefixed class name:

   $a = new fooSubnamespacefoo();
   

   Or,

   fooSubnamespacefoo::staticmethod();
   

   If the present namespace is foonamespace, it will always resolve to foonamespacefooSubnamespacefoo. If the code is global, non-namespaced code, this resolves to fooSubnamespacefoo.

3. Fully qualified name/Prefixed name with global prefix operator:

   $a = new foonamespacefoo();
   

   Or,

   foonamespacefoo::staticmethod();
   

   This always resolves to the literal name laid out in the code, foonamespacefoo.

Now suppose you define a class & call the method of a class within the same namespace.

<?php
    class foo {
        function barFn() {
          echo "Hello foo!"
        }
        function bar() {
            barFn();
            // interpreter is confused which instance's function is called
            $this->barFn();
        }
    }
    $a = new foo();
    $a->bar();
?>

$this pseudo-variable has the methods and properties of the current object. Such a thing is beneficial because it allows you to access all the member variables and methods of the class. Inside the class, it is called $this->functionName(). Outside of the class, it is called $theclass->functionName().

$this is a reference to a PHP object the interpreter created for you, which contains an array of variables. If you call $this inside a normal method in a normal class, $this returns the object to which this method belongs.

Steps to Resolve the Error of Calling to Undefined Function in PHP