User Authentication in Unit Testing Laravel 5.1

Question

I'm a bit confused about the unit testing in Laravel 5.1.

I want to test the updating of settings from a user account. I'm using phpunit for testing.

This is my test case.

/** @test */
    public function it_updates_personal_details(){

        $this->withoutMiddleware();
        $this->visit('/account/settings')
            ->post('/account/settings', ["firstname"=>"RingoUpdated", 'lastname'=>"RingyUpdated", "username"=>"ringo", "bio"=>"My Personal Information", "facebook"=>"myFbUsername", "twitter"=>"myTwUsername", ])
            ->seeJson([
                "signup"=>true
            ]);
    }

But in my controller, I'm using Auth::id() to get the current logged in user's id. How can I mock this in Laravel 5.1 ?

samlev · Accepted Answer · 2015-11-04 21:47:28Z

7

The simplest way is to make use of Mockery in the Facade:

public function it_updates_personal_details(){
    Auth::shouldReceive('id')->andReturn(1);

    $this->withoutMiddleware();
    $this->visit('/account/settings')
        ->post('/account/settings', ["firstname"=>"RingoUpdated", 'lastname'=>"RingyUpdated", "username"=>"ringo", "bio"=>"My Personal Information", "facebook"=>"myFbUsername", "twitter"=>"myTwUsername", ])
        ->seeJson([
            "signup"=>true
        ]);
}

answered Nov 4, 2015 at 21:47

samlev

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6 Comments

user1012181 Over a year ago

When running the test, I'm creating a test user. Is there any way to get the id of that user?

samlev Over a year ago

You could do something like this (it's ugly, but possible): Auth::shouldReceive('id')->andReturn(DB::table('users')->orderBy('id', 'desc')->first(['id'])->id); - keep in mind that I haven't tested this, but that should get the ID of the last created user.

samlev Over a year ago

Actually... that may not work. It would get the last user before the new one you added.

user1012181 Over a year ago

Would there be any other way for doing this?

samlev Over a year ago

Hard to say without knowing your architecture or the function that you're calling. You could use that method, but add 1 to the result, and pretend that that's correct (even less ideal). It really depends on how you're actually creating the user, etc.

|

Pratik Mehta · Accepted Answer · 2018-03-22 07:31:14Z

3

That’s awesome, we can move to class TestCase for all test case and we can give a name, like this:

public function loginWithFakeUser()
{
      $user = new User([
        'id' => 1,
        'name' => 'yish'
      ]);

      $this->be($user);
}

When we need to authenticated user we can call loginWithFakeUser to pretend user.

answered Mar 22, 2018 at 7:31

Pratik Mehta

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1 Comment

Bird87 ZA Over a year ago

Using this, I get a Call to a member function guard() on null at this line: $this->app['auth']->guard($driver)->setUser($user);. This is in Laravel 5.8

cmac · Accepted Answer · 2016-11-17 22:46:11Z

The actingAs helper method. Laravel 5.3

<?php

use Illuminate\Foundation\Testing\WithoutMiddleware;
use Illuminate\Foundation\Testing\DatabaseMigrations;
use Illuminate\Foundation\Testing\DatabaseTransactions;

use App\User; // user model

class ExampleTest extends TestCase
{
    public function testApplication()
    {
        $user = User::find(1); // find specific user

        $this->actingAs($user)
             ->visit('/')
             ->see('Hello, '.$user->name);
    }
}

The actingAs helper method. provides a simple way to authenticate a given user as the current user.

You may also use a model factory to generate and authenticate a user:

        $user = factory(App\User::class)->create();
        $this->actingAs($user)
         ->visit('/')
         ->see('Hello, '.$user->name);

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