Php 如何在其他函数中调用在设置函数中创建的用户
这是我的完整代码,我写它来测试控制器的功能Php 如何在其他函数中调用在设置函数中创建的用户,php,unit-testing,laravel-5,phpunit,laravel-5.1,Php,Unit Testing,Laravel 5,Phpunit,Laravel 5.1,这是我的完整代码,我写它来测试控制器的功能 <?php use Illuminate\Foundation\Testing\WithoutMiddleware; use Illuminate\Foundation\Testing\DatabaseMigrations; use Illuminate\Foundation\Testing\DatabaseTransactions; use Illuminate\Foundation\Testing\Concerns\Makes
<?php
use Illuminate\Foundation\Testing\WithoutMiddleware;
use Illuminate\Foundation\Testing\DatabaseMigrations;
use Illuminate\Foundation\Testing\DatabaseTransactions;
use Illuminate\Foundation\Testing\Concerns\MakesHttpRequests;
class RouteForecastTest extends TestCase
{
protected $password = 'rightpassword';
public function setUp()
{
parent::setUp();
$this->createApplication();
$this->user = factory(App\User::class)->create([
'email' => 'testuser@email.com',
'password' => bcrypt($this->password),
'type' => 'admin'
]);
}
/** @test */
public function createTest()
{
$this->actingAs($user);
$this->visit('/client/ocp/profile/247')
->click('New ')
->seePageIs('/client/ocp/profile/247/create')
->see('name');
}
public function tearDown(){
parent::tearDown();
$this->user->delete();
}
}
如何修复此错误?复制某人的库时,复制所有文件或以不需要的方式重写它
此
尝试引用另一个php中定义的属性用户
。另外,给我们一部分代码可能没有帮助(最后一个花括号和行列表表示代码不完整)。我添加了完整的代码:)可以吗?TestCase
类是否存在,User
类是否存在于文件夹层次结构中它应该位于的位置?我不知道你想要什么,try$this->actingAs($this->User);
ErrorException: Undefined variable: user