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数据库连接后-

`$tbl_name="mytable";`
 $adjacents = 3;
 $query = "SELECT COUNT (*) as num FROM $mytable";
 $total_pages = mysql_fetch_array(mysql_query($query));  this is line 32
 $total_pages = $total_pages[num];
 $targetpage = "pagination.php";  (the name of this file)
 $limit = 20;

我得到的错误是-

警告:mysql_fetch_array():提供的参数不是第 32 行中的有效 MySQL 结果资源。

谁能帮忙?

谢谢

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1 回答 1

1

问题应该在您的查询中,它应该是:

 $query = "SELECT COUNT (*) as num FROM mytable";

或者

 $query = "SELECT COUNT (*) as num FROM ".$tbl_name."";

您正在引用以前未定义的变量 $mytable

于 2012-04-11T07:15:31.713 回答