laravel 4嘲笑模拟模型关系

Question

说我有两个模型，从Eloquent延伸，它们相互关联。我可以嘲笑关系吗？

即：

class Track extends Eloquent { 
    public function courses() 
    { 
     return $this->hasMany('Course'); 
    } 
} 

class Course extends Eloquent { 
    public function track() 
    { 
     return $this->belongsTo('Track'); 
    } 
} 

在MyTest的，我想通过调用跟踪属性，而不是赛道实例（我不创造过程的模拟，并返回轨道，的一个实例要查询生成器）

use \Mockery as m; 

class MyTest extends TestCase { 
    public function setUp() 
    { 
     $track = new Track(array('title' => 'foo')); 
     $course = m::mock('Course[track]', array('track' => $track)); 

     $track = $course->track // <-- This should return my track object 
    } 
} 

Answer 1

由于赛道是一个属性，而不是一个方法，创造了模拟时，你会需要重写setAttribute和getAttribute方法s的模型。下面是一个解决方案，可以让你设置该属性的期望，你正在寻找：

$track = new Track(array('title' => 'foo')); 
$course = m::mock('Course[setAttribute,getAttribute]'); 
// You don't really care what's returned from setAttribute 
$course->shouldReceive('setAttribute'); 
// But tell getAttribute to return $track whenever 'track' is passed in 
$course->shouldReceive('getAttribute')->with('track')->andReturn($track); 

嘲讽Course对象时，你并不需要指定track方法，除非你也想测试代码依赖查询生成器。如果是这种情况，那么你可以嘲笑这样的track方法：

// This is just a bare mock object that will return your track back 
// whenever you ask for anything. Replace 'get' with whatever method 
// your code uses to access the relationship (e.g. 'first') 
$relationship = m::mock(); 
$relationship->shouldReceive('get')->andReturn([ $track ]); 

$course = m::mock('Course[track]'); 
$course->shouldReceive('track')->andReturn($relationship);