問題與saveAssociated cakephp

Question

我有一些問題，試圖保存有4張圖片的文章。事情是，我需要使用文章ID爲了命名圖片像article_id."-"$i

由於我只有4張圖片每篇文章這$我應該從1到4或從0到三。

現在的問題是，爲了實現這一目標，我需要創建和保存文章的模型，所以我可以有一個id使用，但隨後進行所有的腳本，使拇指和形成名字，當我去後文章 - > saveAssociated（）我有兩倍的文章記錄創建！我試圖在保存之前將ID設置爲「-1」，但沒有任何內容...

任何建議將非常感謝！

代碼：

public function add() { 

    if ($this->request->is ('ajax')) { 
     $this->layout = 'ajax'; 
    } else { 
     $this->layout = 'default'; 
    } 



    if ($this->request->is ('post')) { 

     $this->Article->create(); 

     $this->request->data ['Article'] ['time_stamp'] = date ('Y-m-d H:i:s', time()); 




     if ($this->Article->save($this->request->data)) { 


     for ($i=0; $i<4; $i++){ 



      $img_path = "./images/"; 
      $extension[$i] = end(explode('.', $this->request->data['Image'][$i]['image']['name'])); 

      $this->request->data['Image'][$i]['image'] = array('name'=>$this->Article->id."-".$i, 'tmp_name' => $this->request->data['Image'][$i]['image']['tmp_name']); 
     // $this->request->data['Image'][$i]['name'] = $this->Article->id."-".$i; 


      $this->request->data['Image'][$i]['ext']= $extension[$i]; 

      $target_path[$i] = $img_path . basename($this->request->data['Image'][$i]['image']['name'].".".$extension[$i]); 


      if(!move_uploaded_file($this->request->data['Image'][$i]['image']['tmp_name'], $target_path[$i])) { 

       die(__ ('Fatal error, we are all going to die.')); 

      }else{ 

       $this->Resize->img($target_path[$i]); 
       $this->Resize->setNewImage($img_path.basename($this->request->data['Image'][$i]['image']['name']."t.".$extension[$i])); 
       $this->Resize->setProportionalFlag('H'); 
       $this->Resize->setProportional(1); 
       $this->Resize->setNewSize(90, 90); 
       $this->Resize->make(); 

      } 
      } 


      $this->Article->id; 
      pr($this->Article->id); 
      $this->Article->saveAssociated($this->request->data, array('deep' => true)); 
      //$this->redirect (array ('action' => 'view', $this->Article->id)); 
      pr($this->Article->id); 
      exit; 
      $this->Session->setFlash (__ ('Article "' . $this->request->data ["Article"] ["name"] . '" has been saved')); 


     } else { 
      $this->Session->setFlash (__ ('The article could not be saved. Please, try again.')); 
     } 

} 



    $items = $this->Article->Item->find ('list'); 
    $payments = $this->Article->Payment->find ('list'); 
    $shippings = $this->Article->Shipping->find ('list'); 

    $this->set (compact ('items', 'payments', 'shippings')); 
} 

Answer 1

而不是

$this->Article->saveAssociated(); 

這將再次保存的物品，只需保存圖像分別使用這樣的事情：

foreach($this->request->data['Image'] as &$image) { 
    $image['name'] = 'whatever_you_want' . $this->Article->id; 
    $image['article_id'] = $this->Article->id; 
} 
$this->Article->Image->save($this->request->data['Image']); 

另一種選擇（不一定更好 - 只是另一種選擇）而已將新創建的文章的id添加到現有的Article數組，然後saveAssociated()。如果該文章的數據中包含id，則它將更新而不是創建。我建議第一個答案的上方，但是 - 只是集思廣益其他選擇的情況下，這可以幫助某人的情景：

// 1) save the Article and get it's id 
// 2) append the `id` into the Article array 
// 3) do your image-name manipulation using the id 
// 4) saveAssociated(), which updates the Article and creates the Images