Laravel Eloquent 模型单元测试

Question

我正在尝试编写一个测试用例来测试 Laravel 4.2

中两个 Eloquent 模型之间关系的关联和分离

这是我的测试用例：

class BookingStatusSchemaTest extends TestCase
{

  private $statusText = "Confirmed";
  private $bookingStub;
  private $statusStub;

  public function testMigrateService()
  {

    $this->createTestData();

    $booking = $this->bookingStub;
    $status = $this->statusStub;

    /**
     * Check that the booking has no status. OK
     */
    $this->assertNull($booking->status);

    /**
     * Check that status has no booking. OK
     */
    $this->assertEquals(count($status->bookings), 0);

    /**
     * Add a status to the booking. OK
     */
    $booking->status()->associate($this->statusStub);

    /**
     * Check that status has a booking. NOT OK - This gives error
     */
    $this->assertEquals(count($status->bookings), 1);

    /**
     * Check that the booking has a status. OK
     */
    $this->assertNotNull($booking->status);

    /**
     * Do NOT delete the status, just set the reference
     * to it to null.
     */
    $booking->status = null;

    /**
     * And check again. OK
     */
    $this->assertNull($booking->status);
  }

  private function createTestData()
  {

    $bookingStatus = BookingStatus::create([ 
        'status' => $this->statusText 
    ]);

    $booking = Booking::create([ ]);

    $this->bookingStub = $booking;
    $this->statusStub = $bookingStatus;

  }

}

当我执行它时，我得到：

There was 1 failure:

1) BookingStatusSchemaTest::testMigrateService
Failed asserting that 1 matches expected 0.

预订模式：

class Booking extends Eloquent {

  /**
  * A booking have a status
  */
  public function status()
  {
    return $this->belongsTo('BookingStatus');
  }

}

预订状态模型：

class BookingStatus extends Eloquent
{
  protected $table = 'booking_statuses';
  protected $guarded = [ 'id' ];
  protected $fillable = ['status'];

  /**
   * A booking status belongs to a booking
   */
  public function bookings()
  {
    return $this->hasMany('Booking');
  }

}

这是 bookingstatus 的迁移架构：

  Schema::create('booking_statuses', function(Blueprint $table)
  {
    $table->increments('id');
    $table->string('status');
    $table->timestamps();
  });

这里是预订：

Schema::create('bookings', function(Blueprint $table)
{
  $table->increments('id');
  $table->unsignedInteger('booking_status_id')->nullable();
  $table->timestamps();
});

我必须添加/更改什么才能验证我的测试用例中的关系？

Answer 1

It's been a while and I had totally forgotten about this question. Since OP still sems interested in it, I'll try to answer the question in some way.

所以我假设实际任务是：如何测试两个 Eloquent 模型之间的正确关系？

我认为是 Adam Wathan 首先建议放弃像 "Unit Tests" 和 "Functional Tests" 和 "I-have-no-idea-what-this-means Tests" 这样的术语，而只是将测试分成两个 concerns/concepts：特征和单元，其中 Features 简单地描述了应用程序的功能，例如 "A logged in user can book a flight ticket"，而 Units 描述了它的较低级别的单元及其公开的功能，例如 "A booking has a status".

我非常喜欢这种方法，考虑到这一点，我想重构您的测试：

class BookingStatusSchemaTest extends TestCase
{
    /** @test */
    public function a_booking_has_a_status()
    {
        // Create the world: there is a booking with an associated status
        $bookingStatus = BookingStatus::create(['status' => 'confirmed']);
        $booking = Booking::create(['booking_status_id' => $bookingStatus->id]);

        // Act: get the status of a booking
        $actualStatus = $booking->status;

        // Assert: Is the status I got the one I expected to get?
        $this->assertEquals($actualStatus->id, $bookingStatus->id);
    }


    /** @test */    
    public function the_status_of_a_booking_can_be_revoked()
    {
        // Create the world: there is a booking with an associated status
        $bookingStatus = BookingStatus::create(['status' => 'confirmed']);
        $booking = Booking::create(['booking_status_id' => $bookingStatus->id]);

        // Act: Revoke the status of a booking, e.g. set it to null
        $booking->revokeStatus();

        // Assert: The Status should be null now
        $this->assertNull($booking->status);
    }
}

此代码未经测试！

请注意函数名称读起来像是对预订及其功能的描述。您并不真正关心实现，您不必知道 Booking 在哪里或如何获得其 BookingStatus - 您只是想确保如果有带有 BookingStatus 的 Booking，您可以获得该 BookingStatus。或者撤销它。或者也许改变它。或者做任何事情。您的测试显示了您希望如何与该单元交互。所以写测试然后努力让它通过。

您测试中的主要缺陷可能是您有点 "afraid" 会发生一些魔法。相反，将您的模型视为普通的旧 PHP 对象 - 因为它们就是这样！而且你不会运行在 POPO 上进行这样的测试：

/**
 * Do NOT delete the status, just set the reference
 * to it to null.
 */
$booking->status = null;

/**
 * And check again. OK
 */
$this->assertNull($booking->status);

这是一个非常广泛的话题，关于它的每一个陈述都不可避免地带有偏见。有一些准则可以帮助您相处，例如 "only test your own code"，但要将所有和平放在一起真的很难。幸运的是，前面提到的 Adam Wathan 有一个非常出色的视频课程，名为“Test Driven Laravel”，他在其中试驾了整个真实世界的 Laravel 应用程序。它可能有点贵，但它值得每一分钱，并且比 Whosebug 上的一些随机家伙更能帮助您理解测试方式 :)

Answer 2

要测试您是否设置了正确的 Eloquent 关系，您必须运行断言关系 class ($model->relation())。你可以断言

通过断言 $model->relation() 是 HasMany、BelongsTo、HasManyThrough...等的实例，这是正确的关系类型
使用$model->relation()->getRelated()
它使用正确的外键 $model->relation()->getForeignKey()
外键通过使用 Schema::getColumListing($table) 作为列存在于 table 中（这里，如果是 HasMany 关系，$table 要么是 $model->relation()->getRelated()->getTable()或者 $model->relation()->getParent()->getTable() 如果是 BelongsTo 关系）

例如。假设你有一个 Parent 和一个 Child 模型，其中一个 Parent 通过 children() 方法使用 parent_id 作为外来的有许多 Child钥匙。 Parent 映射 parents table 和 Child 映射 children table.

$parent = new Parent;
# App\Parent
$parent->children()
# Illuminate\Database\Eloquent\Relations\HasMany
$parent->children()->getRelated()
# App\Child
$parent->children()->getForeignKey()
# 'parent_id'
$parent->children()->getRelated()->getTable()
# 'children'
Schema::getColumnListing($parent->children()->getRelated()->getTable())
# ['id', 'parent_id', 'col1', 'col2', ...]

编辑此外，这不会触及数据库，因为我们从不保存任何东西。但是，需要迁移数据库，否则模型将不会与任何 table 关联。

Laravel Eloquent 模型单元测试

Laravel Eloquent Model Unit testing

phpunit

laravel

eloquent

laravel-4