mysql_connect("localhost","root","2159645") or die("connect error!"); mysql_select_db("yonghu") or die("select db error"); /*检查该用户是否存在*/ $query="SELECT `password` FROM `xinxi` where `username`='".$_POST['username']."'"; $result=mysql_query($query); $num=mysql_num_rows($result); if($num==0) { echo "您没有注册,请注册后使用!"; exit; } $user=mysql_fetch_array($result, MYSQL_ASSOC); /*检查该用户密码是否正确*/ if($user['password']!=$_POST['password']){ echo "您的密码有误,请再试试。"; exit; } <form method="POST" name="frmlogin" target="_top" action="login.php">
<?php if(! isset($_POST[username])) return; mysql_connect("localhost","root","2159645","yonghu"); /*检查该用户是否存在*/ $query="SELECT username FROM xinxi where username='$_POST[username]'"; $result=mysql_db_query("yonghu",$query); ....
$query="SELECT username FROM xinxi where username='$_POST[username]'"; 改成: $query="SELECT username FROM xinxi where username='{$_POST[username]}'";下面那条SQL语句改法一样的
mysql_connect("localhost","root","2159645") or die("connect error!"); mysql_select_db("yonghu") or die("select db error"); 排除错误的方法: 进行echo 比如: mysql_connect("localhost","root","2159645","yonghu"); echo "ok";//若打印出来 说明上面的连接的对的否则... $query="SELECT username FROM xinxi where username='$_POST[username]'"; echo "ok";//若打印出来,说明上面的也是对的 否则...... $result=mysql_db_query("yonghu",$query); echo "ok";//若打印出来,说明上面的也是对的 否则......自己试着解决,找出错误的根源。 你的SQL语句太不规范了 要加强学习....
mysql_select_db("yonghu");
/*检查该用户是否存在*/
$query="SELECT `username` FROM `xinxi` where `username`='".$_POST['username']."'";
$result=mysql_query($query);
$num=mysql_num_rows($result);
if($num==0)
{
echo "您没有注册,请注册后使用!";
exit;
}
$user=mysql_fetch_array($result, MYSQL_ASSOC);
/*检查该用户密码是否正确*/
if($user['password']!=$_POST['password']){
echo "您的密码有误,请再试试。";
exit;
}
mysql_connect("localhost","root","2159645") or die("connect error!");
mysql_select_db("yonghu") or die("select db error");
/*检查该用户是否存在*/
$query="SELECT `password` FROM `xinxi` where `username`='".$_POST['username']."'";
$result=mysql_query($query);
$num=mysql_num_rows($result);
if($num==0)
{
echo "您没有注册,请注册后使用!";
exit;
}
$user=mysql_fetch_array($result, MYSQL_ASSOC);
/*检查该用户密码是否正确*/
if($user['password']!=$_POST['password']){
echo "您的密码有误,请再试试。";
exit;
}
<form method="POST" name="frmlogin" target="_top" action="login.php">
if(! isset($_POST[username])) return;
mysql_connect("localhost","root","2159645","yonghu");
/*检查该用户是否存在*/
$query="SELECT username FROM xinxi where username='$_POST[username]'";
$result=mysql_db_query("yonghu",$query);
....
改成:
$query="SELECT username FROM xinxi where username='{$_POST[username]}'";下面那条SQL语句改法一样的
mysql_select_db("yonghu") or die("select db error");
排除错误的方法:
进行echo 比如:
mysql_connect("localhost","root","2159645","yonghu");
echo "ok";//若打印出来 说明上面的连接的对的否则...
$query="SELECT username FROM xinxi where username='$_POST[username]'";
echo "ok";//若打印出来,说明上面的也是对的 否则......
$result=mysql_db_query("yonghu",$query);
echo "ok";//若打印出来,说明上面的也是对的 否则......自己试着解决,找出错误的根源。 你的SQL语句太不规范了 要加强学习....