PHP+MYSQL
1.数据库名:login 表名:exam2
字段名 :name password (一共两个)
3.表单里输入框的名字分别为:name password
4.最关键了,login.php页的代码,要求:将用户在表单里填写的name以及password 与 MYSQL数据库中exam表中的name以及password分别进行对比,如果一致,就提示登录成功,否则,弹出对话框:密码或用户名不正确!
第1个回答 2012-11-18
软件？直接抓抓就行
第2个回答 2012-11-18
if(isset($_POST['sub']))
{
$uname=$_POST['name'];
$password=$_POST['pass'];
$sql="select * from `tb_user` where `uname`='".mysql_real_escape_string($uname)."' and `password`='".mysql_real_escape_string$pwd)."'";
$query=mysql_query($sql);
$rs=mysql_fetch_array($query);
$num=mysql_num_rows($query);
if($num==1)
{
header('location:index.php');
$_SESSION['ID'] = $rs['id'];
$_SESSION['uname'] = $rs['uname'];
$_SESSION['role'] = $rs['rid'];
$_SESSION['time'] = date('Y-m-d h:i:s');
}
else
{
echo "<script>alert('密码或用户名不正确!')</script>";
}
}
{
$uname=$_POST['name'];
$password=$_POST['pass'];
$sql="select * from `tb_user` where `uname`='".mysql_real_escape_string($uname)."' and `password`='".mysql_real_escape_string$pwd)."'";
$query=mysql_query($sql);
$rs=mysql_fetch_array($query);
$num=mysql_num_rows($query);
if($num==1)
{
header('location:index.php');
$_SESSION['ID'] = $rs['id'];
$_SESSION['uname'] = $rs['uname'];
$_SESSION['role'] = $rs['rid'];
$_SESSION['time'] = date('Y-m-d h:i:s');
}
else
{
echo "<script>alert('密码或用户名不正确!')</script>";
}
}
