嘲弄與andReturn

Question

返回新對象，我有以下方法：

public function getThreads() 
{ 
    $results = \DB::select(\DB::raw(config('sql.messages.threads')), [$this->user->user_id]); 

    $messagesThreadsModel = app()->make(MessagesThreadsModel::class); 
    $messages = []; 

    foreach ($results as $r) { 
     $message = $messagesThreadsModel->find($r->messages_thread_id); 
     $message->unread = $r->unread; 
     $messages[] = $message; 
    } 

    return $messages; 
} 

爲了測試上面我嘲笑通話\ DB ::選擇（通過Laravel的門面）返回的對象列表如本數據庫類通常做。然後加載消息線程模型，該模型在容器中被再次模擬並替換（因此app（） - > make（）將返回它的模擬實例，而不是實際模型）。

最後這一點：

$messagesThreadsModel->find($r->messages_thread_id); 

再次嘲笑返回一個虛擬對象（存根？）。全部如下：

$threadsList = $this->mockThreads(); 

    // mock the raw expression, check the query 
    \DB::shouldReceive('raw')->once()->with(m::on(function($sql) 
    { 
     return strpos($sql, 'messages') !== false; 
    }))->andReturn(true); 

    // mock the DB call, return a list of objects 
    \DB::shouldReceive('select')->once()->with(true, [$this->usersModel->user_id, $this->usersModel->user_id, $this->usersModel->user_id])->andReturn($threadsList); 

    $mockThreadResult = new \StdClass; 
    $mockThreadResult->last = "date"; 

    $this->messagesThreadModel->makePartial(); 

    // HERE IS THE TRICKY PART! 
    $this->messagesThreadModel->shouldReceive('find')->times(count($threadsList))->andReturn($mockThreadResult); 

    $this->app->instance('App\Freemiom\Models\Messages\MessagesThreads', $this->messagesThreadModel); 

    $messages = new Messages($this->usersModel); 
    $threadList = $messages->getThreads(); 

現在是什麼問題？因爲我傳遞了已創建的虛擬對象，每次在循環中調用 - > find方法時，都會返回相同的對象。

我應該如何告訴嘲笑每次調用返回一個新的對象？它甚至有可能嗎？或者，也許我應該做一些代碼重構，使其全部可測試？

Answer 1

所以爲了能夠返回一個新的對象具有相同的模擬的方法每個連續的電話，我只好用andReturnUsing像這樣：

$this->messagesThreadModel->shouldReceive('find')->times(count($threadsList))->andReturnUsing(function() 
{ 
    $mockThreadResult = new \StdClass; 
    $mockThreadResult->last = "date"; 

    return $mockThreadResult; 
}); 

這會模仿口才模型的行爲也用find（）方法返回新的對象。