以下是我的代码:
<?php
$name=$_POST[$ad_name];
$password=$_POST[$ad_password]; mysql_connect("localhost","root","ps") or
die("Could not connect:".mysql.error());
mysql_select_db("library");
$result=mysql_query("SELECT * FROM administrator WHERE ad_name='$name' and
ad_password='$password'");
$numrow=mysql_num_rows($result);
if($numrow<>0){
echo '<meta HTTP-EQUIV="Refresh" Content="0;URL=/welcome.html">';// 跳到管理员可以处理的界面
}
else{
echo '<meta HTTP-EQUIV="Refresh" Contentf="0;URL=/login.php">';
}
mysql_close();
?>
<body>
<div align="center" class="STYLE1">
<p> </p>
<p>管理员登陆</p>
</div>
<form name="form1" method="post" action="">
<p align="center"> </p>
<p align="center"> </p>
<p align="center"><span class="STYLE2"><strong>管理员姓名</strong>:</span>
<input type="text" name="ad_name" />
</p>
<p align="center"> </p>
<p align="center"><span class="STYLE3 STYLE2">密码:</span>
<input type="text" name="ad_passwort" />
</p>
<p align="center"> </p>
<p align="center"><input type="submit" name="Submit" value="确定" /><input type="reset" name="Submit2" value="重置" /> </p>
</form>
</body>
</html>
一直跳转不来,求问
<?php
$name=$_POST[$ad_name];
$password=$_POST[$ad_password]; mysql_connect("localhost","root","ps") or
die("Could not connect:".mysql.error());
mysql_select_db("library");
$result=mysql_query("SELECT * FROM administrator WHERE ad_name='$name' and
ad_password='$password'");
$numrow=mysql_num_rows($result);
if($numrow<>0){
echo '<meta HTTP-EQUIV="Refresh" Content="0;URL=/welcome.html">';// 跳到管理员可以处理的界面
}
else{
echo '<meta HTTP-EQUIV="Refresh" Contentf="0;URL=/login.php">';
}
mysql_close();
?>
<body>
<div align="center" class="STYLE1">
<p> </p>
<p>管理员登陆</p>
</div>
<form name="form1" method="post" action="">
<p align="center"> </p>
<p align="center"> </p>
<p align="center"><span class="STYLE2"><strong>管理员姓名</strong>:</span>
<input type="text" name="ad_name" />
</p>
<p align="center"> </p>
<p align="center"><span class="STYLE3 STYLE2">密码:</span>
<input type="text" name="ad_passwort" />
</p>
<p align="center"> </p>
<p align="center"><input type="submit" name="Submit" value="确定" /><input type="reset" name="Submit2" value="重置" /> </p>
</form>
</body>
</html>
一直跳转不来,求问
$password=$_POST[$ad_password];
这个有问题,改成$_POST[ad_name],下同,这样子导致搜索结果都为空,所以一直是登陆页,跳转代码光看看没啥问题,先改掉这个再说
另外,最好html代码<html><head>补充完整吧
if ( $_SERVER["REQUEST_METHOD"]=="POST" ){}
$password=$_POST[$ad_password];
将POST里面的改成['ad_name'],另外一个相同如果还不行则将POST改成_REQUEST[]
$result=mysql_query("SELECT * FROM administrator WHERE ad_name='$name' and
ad_password='$password'");這個語句不對吧??$result=mysql_query("SELECT * FROM administrator WHERE ad_name='".$name."' and
ad_password='".$password."'");
if($_POST['Submit']){
$name=$_POST['ad_name'];
$password=$_POST['ad_password']; .....................
mysql_close();
}
?>
...........
ad_password='$password'");
还有你的密码是明文密码吗??
ad_password='$password'");这种写法,很容易被注入。另:
$name=$_POST[$ad_name];
$password=$_POST[$ad_password];
这两个,$name和$password会得不到正确的值,应该是:
$name=$_POST["表单中相应input的name"];
$password=$_POST[{"表单中相应input的name"];