急!!在线等!用PHP如何与数据库进行判断写登录界面!

我只写出一段较为简单的代码~~其他的楼主自己看着办吧.....//这两个是变量,我采用的是post方法获取变量值,楼主可以选择其他方式
$name=$_POST[$username];
$pasw=$_POST[$password];$sql="select * from user where Username='$name' and Password='$pasw'";
$rs=mysql_query($sql);
$reccount=mysql_num_rows($rs);
if($reccount<>0){
header("Location: index.php");//这里是跳去你的网站主页或其他你想跳去的页面
}以上只供参考~~当然这几个语句并不是最好的...可能存在问题~如果有问题大家可以提出来...楼主可以根据我给的大概思路来实现...登录界面很容易实现的..

解决方案 »

免费领取超大流量手机卡，每月29元包185G流量+100分钟通话, 中国电信官方发货

$sql = "select * from tab_name where Username = '".trim($_POST['username']."'");$re = mysql_query($sql);$r = @extract(mysql_fetch_array($re));if ($_POST['Password'] && $_POST['Password'] == $password) {
登陆成功
} else  {
登陆失败
}思路就这样的
这是我写的代码!输入以后没有反应!请各位提一下那里错误!
<?php
$conn = new com("adodb.connection");
$connstr="Provider=OraOLEDB.Oracle.1;Password=SZS;Persist Security Info=True;User ID=SZS;Data Source= oraDP;Extended Properties=''";
$conn->open($connstr);
?><?php
$sql="select * from SZSUser";
$rs=new com("adodb.recordset");
$rs->open($sql,$conn,1,3);
?>
<?php
if($_POST['login']){
  $sql="select * from SZSUser where Username={$_POST['Username']} and Password={$_POST['Password']}";
  $userInfo = &$_conn->GetRow($sql); //adodb
  if($userInfo){
     Header( "Location:./caky.php" );
exit;
   }else{
     Header( "Location:./login.php" );
  }
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
<title>登录界面</title>
<style type="text/css">

</style>
</head><body>
<form method="post">
<p align="center" class="STYLE2">系统登录</p>
<table width=500 border=1 cellspacing=0 cellpadding=0 bgcolor=#8888ff bordercolor=#448844 align=center><tr><td>用户名称：</td>
<td><INPUT id=txtName name="txtName"   style  ="WIDTH: 200px" tabindex="1"
     ></td></tr><tr><td>用户密码：</td>
<td><INPUT id=txtPwd name="txtPwd" type=password  style  ="WIDTH: 200px" tabindex="2" ></td></tr></table>
<div align="left"></div>
<p align="center">
  <label>
  <input type="submit" name="Submit" style   ="HEIGHT: 24px; WIDTH: 100px" tabindex="2" value="登录" />
  <input type="submit" name="Submit" style   ="HEIGHT: 24px; WIDTH: 100px" tabindex="2" value="取消" />
  </label>
</form>
</body>
</html>
我发先两个问题１：$sql="select * from SZSUser where Username={$_POST['Username']} and Password={$_POST['Password']}";
  应该是　$sql="select * from SZSUser where Username='{$_POST['Username']}' and Password='{$_POST['Password']}'";

2:<INPUT id=txtName name="txtName"  style  ="WIDTH: 200px" tabindex="1" >名字是txtName可是你Username={$_POST['Username']} 先改了再说把
<INPUT id=txtName name="txtName"
{$_POST['Username']}
要相对应,
要么就是{$_POST['txtName']}
密码也一样
<?php
//引用文件
//require_once('./config.inc.php');$conn = new com("adodb.connection");
$connstr="Provider=OraOLEDB.Oracle.1;Password=SZS;Persist Security Info=True;User ID=SZS;Data Source= oraDP;Extended Properties=''";
$conn->open($connstr);$sql="select * from SZSUser";
$rs=new com("adodb.recordset");
$rs->open($sql,$conn,1,3);
if($_POST['login']){
  $sql="select * from SZSUser where Username={$_POST['txtName']} and Password={$_POST['txtPwd']}";
  $userInfo = &$_conn->GetRow($sql); //adodb
  if($userInfo){
     Header( "Location:./caky.php" );
exit;
   }else{
     Header( "Location:./curve.php" );
  }
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
<title>登录界面</title>
<style type="text/css">

</style>
</head><body>
<form method="post">
<p align="center" class="STYLE2">系统登录</p>
<table width=500 border=1 cellspacing=0 cellpadding=0 bgcolor=#8888ff bordercolor=#448844 align=center><tr><td>用户名称：</td>
<td><INPUT id=txtName name="txtName"   style  ="WIDTH: 200px" tabindex="1" ></td></tr><tr><td>用户密码：</td>
<td><INPUT id=txtPwd name="txtPwd" type=password  style  ="WIDTH: 200px" tabindex="2" ></td></tr></table>
<div align="left"></div>
<p align="center">
  <label>
  <input type="submit" name="login" style   ="HEIGHT: 24px; WIDTH: 100px" tabindex="2" value="登录" />
  <input type="submit" name="Submit" style   ="HEIGHT: 24px; WIDTH: 100px" tabindex="2" value="取消" />
  </label>
</form>
</body>
</html>
已经对应!但是他们提示:"Fatal error: Call to a member function GetRow() on a non-object in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\index1.php on line 16
"错误!
很明显啊你new 一个adodb对象是$conn
但是你用的时候又&$_conn->GetRow($sql); //adodb $_conn怎么来的啊？因该是$conn->GetRow把？
楼上的我按照你说的改了,提示:"Fatal error: Uncaught exception 'com_exception' with message '<b>Source:</b> ADODB.Connection<br/><b>Description:</b> 参数类型不正确，或不在可以接受的范围之内，或与其他参数冲突。' in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\index1.php:16 Stack trace: #0 C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\index1.php(16): com->GetRow('select * from S...') #1 {main} thrown in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\index1.php on line 16
"
$sql="select * from SZSUser where Username='{$_POST['txtName']}' and Password='{$_POST['txtPwd']}'";
带上单引号试试。
再不行，echo $sql; 看看sql写的对吗。