MySQL:mysql_fetch_assoc():提供的参数不是有效的 MySQL 结果资源 [英] MySQL: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource

问题描述

可能的重复:
警告:mysql_fetch_array():提供的参数不是有效的 MySQL 结果

我有这个错误:

mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/marlon/domains/webmasterplaats.nl/public_html/edit.php on line 36

这是代码:

    <?php
    $ip = $_SERVER['REMOTE_ADDR'];
    $toegang[] = '86.91.195.26';
    $toegang[] = '84.86.189.70';

    $valid = true;
    if(in_array($ip, $toegang) || isset($valid))
    {
 if(isset($_GET['id']))
 {
  if($_SERVER['REQUEST_METHOD'] == 'POST')
  {
   mysql_query("UPDATE news SET titel='" . mysql_real_escape_string($_POST['titel']) . "', inhoud='" . mysql_real_escape_string($_POST['edit2']) . "' WHERE id='" . mysql_real_escape_string($_GET['id']) . "'");

   echo 'Met success geupdate.' ;
  }
   $database = mysql_connect('localhost','marlonhe19','123456789asd');
   mysql_select_db('wmp', $database);

  $id = $_GET['id'];

  $mysql = mysql_query("SELECT * FROM news WHERE id='$id' ;");

  while($row = mysql_fetch_assoc($mysql)){
   $id = $row['id'];
   $titel = $row['titel'];
   $inhoud = $row['inhoud'];

  echo '
  <form id="form1" name="form1" method="post" action="">
  <input type="text" name="titel" value="$titel" /><br />
  <textarea name="edit2">$inhoud</textarea> <br />
  <input type="submit" name="Submit" value="Opslaan" />';
    }
    }
    }

有什么问题吗?

推荐答案

警告:可能存在 SQL 注入.您的查询似乎失败了.

Warning: SQL injection possible. It looks like your query failed.

替换这个:

$mysql = mysql_query("SELECT * FROM news WHERE id='$id' ;");

与:

$mysql = mysql_query("SELECT * FROM news WHERE id='$id' ;") or die(mysql_error());

你应该自己制作错误处理函数，最好显示错误信息，不要立即退出.

You should make your own error handling function, it's prefferable to display an error message, without exiting immediately.

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