Warning: mysql_num_rows(): supplied argument is not a valid MySQL result res

来源：百度知道 编辑：UC知道 时间：2024/05/31 16:23:00
<?php
$userna = $_POST['username'];
$passwd = $_POST['password'];
$db_name = "shop" ;
$sql = "SELECT * FORM login WHERE username='$userna' " ;
//$query = 'SELECT * FROM user where username="limuyan"';
//连接到服务器
$user = "root" ;
$passwd = "123456" ;
//$connect = mysql_connect('localhost',$user,$passwd)连接到服务器
$connect =mysql_connect('localhost',$user,$passwd) or die('could not connect:'.mysql_error());
//选择数据库
$db = mysql_select_db($db_name);
//执行查询语句
@$result = mysql_query($sql);
//返回记录数mysql_num_rows
$row=mysql_num_rows($result);

if($row==0){
echo "用户名错误" ;
}
else {
if(mysql_result($result,0,"password")==$passwd){
echo "登录成功!" ;
}
else{
echo "密码不正确!" ;
}
}
?

$sql = "SELECT * FORM login WHERE username='$userna' "中的FORM写错了，应该是FROM

$result 不是资源,意思是 query的时候sql可能就错了
把
@$result = mysql_query($sql);
前面的@去掉,错误打印出来,就知道哪里错了

$db = mysql_select_db($db_name);我记得应该是$db = mysql_select_db($db_name,$connection);