PHP警告:mysql_fetch_array():提供的参数不是有效的MySQL结果资源

时间:2013-07-16 15:15:03

标签: php mysql database arrays fetch

将网站上传到我的网络服务器后,我收到以下消息:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource 
in /home1/m1k3ey/public_html/MikeyDev.com/teamdesire/playersheet.php on line 8

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource 
in /home1/m1k3ey/public_html/MikeyDev.com/teamdesire/playersheet.php on line 9

我的PHP版本是:

PHP Version 5.2.17

我无法看到我的代码中出错的地方,任何人都可以协助:

          mysql_select_db('teamdesire', $link);
          $query = "SELECT * FROM playershowercase";
          $result = mysql_query($query,$link);
          $row = array();
Line 8 >  while($row[] = mysql_fetch_array($result));
Line 9 >  $count = mysql_num_rows($result);
          $random = rand(0,$count-1);

3 个答案:

答案 0 :(得分:0)

可能是以下情况?

      mysql_select_db('teamdesire', $link);
      $query = "SELECT * FROM playershowercase";
      $result = mysql_query($query,$link);
      $row = mysql_fetch_assoc($result);
      $count = mysql_num_rows($result);
      $random = rand(0,$count-1);

或者

      mysql_select_db('teamdesire', $link);
      $query = "SELECT * FROM playershowercase";
      $result = mysql_query($query,$link);
      while ($row = mysql_fetch_array($result, MYSQL_NUM))
      $count = mysql_num_rows($result);
      $random = rand(0,$count-1);

答案 1 :(得分:0)

可能的错误来源:

  • Db未在指定的套接字或端口上运行。
  • 对表或数据库的权限不足。
  • 登录凭据不正确。
  • 表格不存在或拼写错误。

排除故障时每行的回音行。

is_resource($link) or die('Could not connect');
mysql_query(...);
echo mysql_error();

答案 2 :(得分:-1)

试试这个:

          mysql_select_db('teamdesire', $link);
          $query = "SELECT * FROM playershowercase";
          $result = mysql_query($query,$link);
          $row = array();
          while($result = mysql_fetch_array($result))
          {
          $row[]=$result;
          }
          $count = mysql_num_rows($result);
          $random = rand(0,$count-1);