2017-08-08 72 views
2

當我運行這段代碼,始終顯示在網頁「結果錯誤」或「沒有數據即時拍攝」知道我添加在數據庫中的數據:其中是錯誤的始終顯示「結果錯誤」

<?php 
if(isset($_POST['update'])) 
      { 
        $data = getpost(); 
        $search_Query = "SELECT * FROM employee WHERE user_id_assets = $data[0]"; 
        $search_result = mysqli_query($conn , $search_Query); 

       if($search_result) 
       { 
        if(mysqli_num_rows($search_result)) 
        { 
         while ($row = mysqli_fetch_array($search_result)) { 
          $user_id_assets = $row['user_id_assets']; 
          $name_emp = $row['name_emp']; 
          $user_account = $row['user_account']; 
          $mail_account = $row['mail_account']; 
        } 
       }else{ 
        echo "no Data For This User ID"; 
       }  
      }else { 
       echo "result Error"; 
      } 
     } ?> 
+0

從這段代碼可以看出'$ conn'從來沒有定義過? –

+0

似乎這個'$ data [0]'是空的/ null –

+0

$ conn mighat已經在連接文件中定義了,代碼是不完整的... –

回答

0

試試這個...

if(isset($_POST['update'])){ 
    $search_Query = 'SELECT * FROM employee WHERE user_id_assets = '.$_POST['user_id_assets']; 
    $search_result = mysqli_query($conn , $search_Query) or mysqli_error($conn); 

    if(mysqli_num_rows($search_result)){ 
     while ($row = mysqli_fetch_array($search_result)) { 
      $user_id_assets = $row['user_id_assets']; 
      $name_emp = $row['name_emp']; 
      $user_account = $row['user_account']; 
      $mail_account = $row['mail_account']; 
     } 
    }else{ 
     echo "no Data For This User ID"; 
    }  
} 
+0

這是結果: 警告:mysqli_num_rows()期望參數1爲mysqli_result,布爾在C:\ xampp \ htdocs \ dashboard \ Website \ process.php中給出60行 否這個用戶ID的數據 –

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什麼'$ data = getpost()'返回? @RaedTattan –

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這是: \t功能的getPost() \t \t \t { \t \t \t \t $交=陣列(); \t \t \t \t $ post [0] = $ _POST ['user_id_assets']; \t \t \t \t $ post [1] = $ _POST ['name_emp']; \t \t \t \t $ post [2] = $ _POST ['user_account']; \t \t \t \t $ post [3] = $ _POST ['mail_account']; \t \t \t \t \t \t \t} –