laravel 4嘲讽模拟模型关系

veilig 发表于 Dev

安全

说我有两个扩展的模型，Eloquent它们彼此关联。我可以嘲笑这种关系吗？

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中，我当然想创建一个模拟，并通过调用track属性而不是track实例返回track的实例（我不希望使用查询构建器）

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
    }
}

杰夫·兰伯特

由于track是属性而不是方法，因此在创建模拟时，您将需要覆盖模型的setAttribute和getAttribute方法。以下是一个解决方案，可让您对要查找的属性设置期望：

$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);

track嘲笑Course对象时，无需指定方法，除非您还想测试依赖于查询生成器的代码。如果是这种情况，则可以模拟如下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);

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