gpt4 book ai didi

PHP登录错误基本

转载 作者:行者123 更新时间:2023-11-29 11:31:35 25 4
gpt4 key购买 nike

我正在 Windows 10 上使用 server @ localhost 运行 xampp Apache 服务器。php 代码不起作用,问题出在哪里?

PS 我正在学习 php,任何帮助将不胜感激

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">

<html>
<head>
<TITLE>Welcome</TITLE>
</head>
<body>
<form action="" method=POST>
<div>
<strong>Username:</strong><input type="text" name="username" value=""><br/>
<strong>Password:</strong><input type="password" name="password" valuse=""><br/>
<input type="submit" name="submit" value="Submit"><br/>
</div>
</forum>
<?php
session_start();
echo "Hello world!";
include('sql_connect.db.php');
if(isset($_POST[submit])){
$uname=$_POST['username']
$pass=$_POST['password']
}
$result = mysql_query("SELECT * FROM users WHERE username=$uname")
$row=mysql_fetch_array($result);
$flag=0
if($row)
$usp=$row['password']
if (pass==usp){
$_SESSION['login_user'] = $uname;
header("location: success.php");
}else {
$error = "Your Login Name or Password is invalid";
}
?>
</body>
</html>

最佳答案

<div class="snippet" data-lang="js" data-hide="false" data-console="true">
<div class="snippet-code">
<pre class="snippet-code-html lang-html prettyprint-override"><code> <html>
<head>
<TITLE>Welcome</TITLE>
</head>
<body>
<form action="checklogin.php" method="POST">
<div>
<strong>Username:</strong><input type="text" name="username"><br/>
<strong>Password:</strong><input type="password" name="password" ><br/>
<input type="submit" name="submit" value="Submit"><br/>
</div>
</form>
</body>
</html></code></pre>
</div>
</div>


<div class="snippet" data-lang="js" data-hide="false" data-console="true">
<div class="snippet-code">
<pre class="snippet-code-html lang-html prettyprint-override"><code> <?php
//code for checklogin.php file
session_start();
echo "Hello world!";
include('sql_connect.db.php');

$uname=$_POST['username'];
$pass=$_POST['password'];

$result = mysql_query("SELECT * FROM users WHERE username=$uname");
$row=mysql_fetch_array($result);
$flag=0;
if($row)
$usp=$row['password'];
if (pass==usp){
$_SESSION['login_user'] = $uname;
header("location: success.php");
}else {
$error = "Your Login Name or Password is invalid";
}
?></code></pre>
</div>
</div>

关于PHP登录错误基本,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37313330/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com