我尝试为我的
User
和Shop
模型之间的关系创建一个单元测试,但是当我运行vendor\\bin\\phpunit
时抛出此错误,我不知道这一点,因为我是单元新手测试。我尝试在控制器上运行我的代码,看看这种关系是否确实有效,幸运的是它按预期工作,但在 phpunit 中运行时却不然。我做错了什么导致这个 phpunit 不能与模型一起使用?
Fatal error:
Uncaught Error: Call to a member function connection() on null in E:\projects\try\vendor\laravel\framework\src\Illuminate\Database\Eloquent\Model.php:1013
Stack trace:
E:\projects\try\vendor\laravel\framework\src\Illuminate\Database\Eloquent\Model.php(979): Illuminate\Database\Eloquent\Model::resolveConnection(NULL)
这是我的 UserTest.php
<?php
namespace Tests\Unit;
use Tests\TestCase;
use Illuminate\Foundation\Testing\DatabaseMigrations;
use Illuminate\Foundation\Testing\DatabaseTransactions;
use App\User;
use App\Shop;
class UserTest extends TestCase
{
protected $user, $shop;
function __construct()
{
$this->setUp();
}
function setUp()
{
$user = new User([
'id' => 1,
'first_name' => 'John',
'last_name' => 'Doe',
'email' => '[email protected]',
'password' => 'secret',
'facebook_id' => null,
'type' => 'customer',
'confirmation_code' => null,
'newsletter_subscription' => 0,
'last_online' => null,
'telephone' => null,
'mobile' => null,
'social_security_id' => null,
'address_1' => null,
'address_2' => null,
'city' => null,
'zip_code' => null,
'signed_agreement' => 0,
'is_email_confirmed' => 0
]);
$user = User::find(1);
$shop = new Shop([
'id' => 1,
'user_id' => $user->id,
'name' => 'PureFoods Hotdog2',
'description' => 'Something that describes this shop',
'url' => null,
'currency' => 'USD'
]);
$user->shops()->save($shop);
$shop = new Shop([
'id' => 2,
'user_id' => $user->id,
'name' => 'PureFoods Hotdog',
'description' => 'Something that describes this shop',
'url' => null,
'currency' => 'USD'
]);
$user->shops()->save($shop);
$this->user = $user;
}
/** @test */
public function a_user_has_an_id(){
$user = User::find(1);
$this->assertEquals(1, $user->id);
}
/** @test */
public function a_user_has_a_first_name()
{
$this->assertEquals("John", $this->user->first_name);
}
/** @test */
public function a_user_can_own_multiple_shops()
{
$shops = User::find(1)->shops()->get();
var_dump($this->shops);
$this->assertCount(2, $shops);
}
}
看来,这个错误是由这行代码引起的:
$user->shops()->save($shop);
- 此代码在我的示例路由或控制器中运行时实际上有效,但在 errors
中运行时抛出
phpunit
用户.php
<?php
namespace App;
use Illuminate\Notifications\Notifiable;
use Illuminate\Foundation\Auth\User as Authenticatable;
class User extends Authenticatable
{
use Notifiable;
protected $guarded = [ 'id' ];
protected $table = "users";
/**
* The attributes that should be hidden for arrays.
*
* @var array
*/
protected $hidden = [
'password', 'remember_token',
];
/**
* returns the Shops that this user owned
*/
public function shops()
{
return $this->hasMany('App\Shop');
}
}
Shop.php
<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
class Shop extends Model
{
protected $guarded = [ 'id' ];
protected $table = "shops";
/**
* returns the Owner of this Shop
*/
public function owner()
{
return $this->belongsTo('App\User');
}
}
任何帮助将非常感激。谢谢!
还有一个原因
检查测试类是否正在扩展 使用 Tests\TestCase; 而不是 使用 PHPUnit\Framework\TestCase;
Laravel 附带了后者,但是 Tests\TestCase 类负责设置应用程序,否则如果模型扩展 PHPunit\Framework\TestCase,则模型将无法与数据库通信。
首先,
setUp()
在每次测试之前都会被调用,所以你不应该在构造函数中调用它
其次,您应该在
parent::setUp()
中调用 setUp()
来初始化应用程序实例。
示例:
<?php
class ExampleTest extends TestCase
{
private $user;
public function setUp()
{
parent::setUp();
$this->user = \App\User::first();
}
public function testExample()
{
$this->assertEquals('[email protected]', $this->user->email);
}
}
解决方案
使用PHPunit\Framework\TestCase
放这个
使用测试\测试用例
问题原因:您无法从 UserTest 类的构造函数中调用的代码中使用 Laravel Models 功能 - 即使您将代码放在“setUp”方法中,您也不必要从构造函数中调用它。 SetUp 由 phpUnit 调用,无需在构造函数中执行。
当 UserTest 构造函数运行时,Laravel Bootstrap 代码尚未被调用。
当调用 UserTest->setUp() 方法时,Laravel Bootstrap 代码已运行,因此您可以使用模型等。
class UserTest extends TestCase
{protected $user, $shop;
function __construct()
{
$this->setUp(); // **THIS IS THE PROBLEM LINE**
}
function setUp()
{
$user = new User([....
当您尝试从 dataprovider 函数读取数据库时,可能会发生这种情况。就像我尝试过的那样,是的。工作方式:
protected static $tariffs_default;
protected function setUp(): void {
parent::setUp ();
if (!static::$tariffs_default){
static::$tariffs_default = DB::...;
}
}
// in dataprovider we use string parm, named as our static vars
public static function provider_prov2(){
return [
[".....", [ 'tariffs'=>'tariffs_default'] ],
];
}
// then, in test we can ask our data by code:
public function testSome(string $inn, string $ogrn, $resp_body){
if ( $ta_var = Arr::get( $resp_body, 'tariffs' ) ){
Arr::set($resp_body, 'tariffs', static::$$ta_var );
}
$this->assert...
}
尝试
composer dump-autoload
~问候
在 PEST PHP 中使用这个:
<?php
use Tests\TestCase;
uses(TestCase::class);
...