2016-12-07 75 views
0

我正在更新圖像文件,但當我更新新的圖像文件時,它會替換舊的圖像文件。我希望保留所有圖像文件而不更換。Codeigniter更新數據庫列中的新圖像,無需替換舊的圖像

例如

if i remove one of them and add new one it update column value with one and remove old one..i need all of them.

如果我刪除其中的一個,並添加新的將其與一個更新列值,並刪除舊one..i需要所有的人。

這裏我上傳代碼:

$picture=array(); 
$cpt = count($_FILES['my_image_upload']['name']); 
$config['upload_path'] = APPPATH.'uploads/'; 
$config['allowed_types'] = '*'; 
$files = $_FILES; 
for($i=0; $i<$cpt; $i++) 
{  
if($_FILES['my_image_upload']['name'][$i]!='') 
    { 
    $_FILES['my_image_upload']['name']= $files['my_image_upload']['name'][$i]; 
    $_FILES['my_image_upload']['type']= $files['my_image_upload']['type'][$i]; 
    $_FILES['my_image_upload']['tmp_name']= $files['my_image_upload']['tmp_name'][$i]; 
    $_FILES['my_image_upload']['error']= $files['my_image_upload']['error'][$i]; 
    $_FILES['my_image_upload']['size']= $files['my_image_upload']['size'][$i]; 
    $this->load->library('upload', $config); 
    $this->upload->initialize($config); 
    $upload= $this->upload->do_upload('my_image_upload'); 
    if($upload) 
     { 
      $this->upload->data(); 
      $picture[] =$_FILES['my_image_upload']['name']; 
     } 
    } 
} 
$fileName = implode('|',$picture); 

在此先感謝..

+0

你如何保存圖像名稱介紹數據庫? –

+0

所有圖像名稱內爆$文件名 – Gulshan

+0

你如何將圖像名稱存儲在數據庫中? –

回答

0

我建議創建新的表像這樣和插入數據...

id user_id image_name 
1  2  abc.jpg 
2  2  bbb.jpg 
3  2  ccc.jpg 
......... 

和簡單的方式插入,更新刪除爲你希望

+0

其工作對我來說..謝謝:) – Gulshan