这是代码:提示的错误行为最后一行。<?php
include 'includes/conn.php';
$sql="select count(*) from user where user='$_POST[user]'and pwd ='$_POST[pwd]'";
$res=mysql_query($sql);
$row=mysql_fetch_array($res);
if($row[0] ==1)
{
echo "< script language='javascript' type = 'text/javascript'>"; echo "window.location.href = 'backgroundPage.php'";
echo "</script>";
}
?>
谢谢各位大神!!php
include 'includes/conn.php';
$sql="select count(*) from user where user='$_POST[user]'and pwd ='$_POST[pwd]'";
$res=mysql_query($sql);
$row=mysql_fetch_array($res);
if($row[0] ==1)
{
echo "< script language='javascript' type = 'text/javascript'>"; echo "window.location.href = 'backgroundPage.php'";
echo "</script>";
}
?>
谢谢各位大神!!php
解决方案 »
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