2017-02-19 33 views
-5

當我想要獲取我的數據庫信息時,發現此錯誤。此行發現錯誤:警告:mysql_fetch_assoc()期望參數1是資源,布爾在第52行給出的C: xampp htdocs xyz adminpanel.php

while($row = mysql_fetch_assoc($result)) 

完整的代碼如下所示,我的代碼中是否有任何錯誤?錯誤顯示期望參數1是資源,我需要修復解決方案。

<html> 
     <head> 
     <link rel="stylesheet" type="text/css" href="css/style.css"> 

      <meta charset="utf-8"> 
      <meta http-equiv="X-UA-Compatible" content="IE=edge"> 
      <meta name="viewport" content="width=device-width, initial-scale=1"> 
      <title>Admin Panel</title> 
      <link rel="stylesheet" href="assets/demo.css"> 
      <link rel="stylesheet" href="assets/form-labels-on-top.css"> 
     </head> 
     <body> 
      <header> 
       <h1 align="center"><b>All User Information</b></h1></br>  
      </header> 


      <?php 
        $servername = "localhost"; 
        $username = "root"; 
        $password = ""; 
        $dbname = "xyz"; 

        // Create connection 
        $conn = new mysqli($servername, $username, $password, $dbname); 
        // Check connection 
        if ($conn->connect_error) 
        { 
         die("Connection failed: " . $conn->connect_error); 
        } 


       //execute the SQL query and return records 
       $result = mysql_query("SELECT * FROM user_information"); 
       //if(!$result) die ('Unable to run query:'.mysql_error()); 
       ?> 
       <table border="2" style= "background-color: #84ed86; color: #761a9b; margin: 0 auto;" > 
       <thead> 
       <tr> 
        <th>User ID</th> 
        <th>User Name</th> 
        <th>User Country</th> 
        <th>User Email</th> 
        <th>User Phone</th> 
        <td>User Address</td> 
        <td>User Password</td> 
       </tr> 
       </thead> 
       <tbody> 
       <?php 
        while($row = mysql_fetch_assoc($result)){ 
        echo 
        "<tr> 
         <td>{$row['user_id']}</td> 
         <td>{$row['user_name']}</td> 
         <td>{$row['user_country']}</td> 
         <td>{$row['user_email']}</td> 
         <td>{$row['user_phone']}</td> 
         <td>{$row['user_address']}</td> 
         <td>{$row['user_password']}</td> 
        </tr>\n"; 
        } 
       ?> 
       </tbody> 
      </table> 




     </body> 
     </html> 

回答

0

替換該行的代碼到代碼2線 $結果= mysqli_query($康恩, 「SELECT * FROM user_information」);

while($row = mysqli_fetch_assoc($result)){ 
+0

是的,我已經做到了。感謝您的反饋意見。 – Nayan

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