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[英]Laravel 5.6 Unit Test Call to a member function beginTransaction() on null
[英]Laravel 5 Unit Test - Call to a member function connection() on null
我嘗試為我的User
和Shop
模型之間的關系創建一個單元測試,但是當我運行vendor\\bin\\phpunit
時會拋出這個錯誤,我對此一無所知,因為我是單元測試的新手. 我試圖在我的控制器上運行我的代碼以查看關系是否真的有效,幸運的是它按預期工作,但在 phpunit 中運行時卻沒有。 我做錯了什么讓這個 phpunit 不能與模型一起工作?
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)
這是我的 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' => 'JohnDoe@example.com',
'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);
}
}
看來,這個錯誤是由這行代碼引起的: $user->shops()->save($shop);
- 此代碼在我的示例路由或控制器中運行時實際上有效,但在phpunit
中運行時拋出errors
用戶.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');
}
}
商店.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');
}
}
任何幫助將不勝感激。 謝謝!
還有一個原因
檢查測試類是否正在擴展使用 Tests\TestCase; 而不是使用 PHPUnit\Framework\TestCase;
Laravel 附帶了后者,但Tests\TestCase類負責設置應用程序,否則如果模型擴展PHPunit\Framework\TestCase ,它們將無法與數據庫通信。
首先, setUp()
在每次測試之前都會被調用,所以你不應該在構造函數中調用它
其次,您應該在 setUp() 中調用parent::setUp()
setUp()
來初始化應用程序實例。
例子:
<?php
class ExampleTest extends TestCase
{
private $user;
public function setUp()
{
parent::setUp();
$this->user = \App\User::first();
}
public function testExample()
{
$this->assertEquals('victor@castrocontreras.com', $this->user->email);
}
}
一個解法
使用 PHPunit\Framework\TestCase
把這個
使用測試\TestCase
當您嘗試從 dataprovider 函數讀取數據庫時,可能會發生這種情況。 像我試過的,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...
}
問題的原因:您不能從在 Class UserTest 的構造函數中調用的代碼中使用 Laravel 模型功能 - 即使您將代碼放在方法“setUp”中,您也不必要地從構造函數中調用它。 SetUp 由 phpUnit 調用,您無需在構造函數中執行此操作。
當 UserTest 構造函數運行時,Laravel Bootstrap 代碼還沒有被調用。
當調用 UserTest->setUp() 方法時,Laravel Bootstrap 代碼已經運行,所以你可以使用你的模型等。
class UserTest extends TestCase
{protected $user, $shop;
function __construct()
{
$this->setUp(); // **THIS IS THE PROBLEM LINE**
}
function setUp()
{
$user = new User([....
嘗試composer dump-autoload
〜問候
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