php:Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

2024-12-19 12:56:28
推荐回答(2个)
回答1:

sql语法错误
$sql = mysql_query("select * from `Out_Check`");
$info = mysql_fetch_array($aql);
改成:
$sql = mysql_query("select * from `Out_Check` where 1");
$info = mysql_fetch_array($sql);

你下面也有错误:
$info = mysql_fect_array($sql); 改成 mysql_fetch_array(); //fetch单词写错

回答2:

$aql
$sql
一个字母之差