Question

我试图显示数据库中所有用户的信息并且不断给我这个错误

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in /home/killerdu/public_html/userlist.php on line 30

第30行

while($people=mysql_fetch_assoc($people_list)){

<?php
include 'core/int.php';
include 'includes/head.php';
include 'head.php';
include 'includes/body.php';
include 'body.php';

$people_list="SELECT * FROM users";

$people=mysql_query($people_list);

?>
<html>
<head>
</head>
<body>
<pre>
<table class="table table-bordered">
      <thead>
        <tr>
          <th>ID</th>
          <th>Username</th>
          <th>Email</th>
        </tr>

      <?php

     while($people=mysql_fetch_assoc($people_list)){ 
     echo "<tr>";
          echo "<td>". $people_list['user_id']."</td>";
          echo "<td>". $people_list['username']."</td>";
           echo "<td>". $people_list['email']."</td>";
        echo "</tr>";     
     }

      ?>
      </thead>
      </pre>
</body>
</html>

Answer 1

根据OP的要求，回答评论：

您混合了while($people=mysql_fetch_assoc($people_list))变量的变量。 while($people_list=mysql_fetch_assoc($people))

Answer 2

您使用了$people=mysql_query($people_list);

所以加载它：

while($row=mysql_fetch_assoc($people)) {
                             // ^ load $people, the resource, not $people_list the string

您加载了查询字符串而不是资源。

警告：mysql_fetch_assoc（）需要参数1

2 个答案: