PHP update table - PHP update table

我正在创建一个用于基本PHP学习目的的虚拟应用程序，现在我的应用程序将用户登录到一个用户可以做多种事情的帐户中，例如搜索产品等。其中一个功能的用户可以做的是编辑帐户的详细信息，这是一切都错了我。没有显示任何错误，但当我通过点击"编辑帐户"按钮提交查询时，一切都可以正常工作，但当我检查数据库时，我发现没有任何改变…

我也知道代码中的安全问题，但现在我想建立基本功能

userEditAccount.php:

<?php

session_start();

    include('connect_mysql.php');

        if(isset($_POST['Edit Account']))
        {
            $usernameNew = stripslashes(mysql_real_escape_string($_POST["username"]));
            $passwordNew = stripslashes(mysql_real_escape_string($_POST["password"]));
            $first_nameNew = stripslashes(mysql_real_escape_string($_POST["first_name"]));
            $last_nameNew = stripslashes(mysql_real_escape_string($_POST["last_name"]));
            $emailNew = stripslashes(mysql_real_escape_string($_POST["email"]));

            $dbusername = $_SESSION['username'];
            $editQuery = mysql_query("UPDATE users SET user_id='NULL' username='$usernameNew', password='$passwordNew', first_name='$first_nameNew', last_name='$last_nameNew' , email='$emailNew' WHERE username='$edit'");

            if(!$editQuery)
            {
                echo mysql_error($editQuery);
                die($editQuery);
            }
        }
?>

<html>
<head>
<title>Edit Account</title>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<link href="style.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="wrapper">
<header><h1>E-Shop</h1></header>
    <article>
        <h1>Welcome</h1>
            <h1>Edit Account</h1>
<div id="login">
    <ul id="login">
    <form method="post" name="editAccount" action="userEditAccount.php"  >
        <fieldset>  
            <legend>Fill in the form</legend>
                <label>Select Username : <input type="text" name="username" /></label>
                <label>Password : <input type="password" name="password" /></label>
                <label>Enter First Name : <input type="text" name="first_name" /></label>
                <label>Enter Last Name : <input type="text" name="last_name" /></label>
                <label>Enter E-mail Address: <input type="text" name="email" /></label>
        </fieldset>
        <br />

                <input type="submit" value="Edit Account" class="button">

    </form>
</div>
    <form action="userhome.php" method="post">
    <div id="login">
        <ul id="login">
            <li>
                <input type="submit" value="back" onclick="index.php" class="button">   
            </li>
        </ul>
    </div>      
</article>
<aside>
</aside>
<div id="footer">Text</div>
</div>
</body>
</html>

我还将包含login.php:

<?php
session_start();
    require('connect_mysql.php');

     if($_SERVER['REQUEST_METHOD'] == 'POST')
     {
        $username = $_POST["username"];
        $password = $_POST["password"];
        $username = stripslashes($username);
        $password = stripslashes($password);
        $username = mysql_real_escape_string($username);
        $password = mysql_real_escape_string($password);
        $query = mysql_query("SELECT * FROM users WHERE Username='$username' AND Password='$password'");
        $numrow = mysql_num_rows($query);
    if($username && $password){
        $query = mysql_query("SELECT * FROM  users WHERE username='$username'");
        $numrow = mysql_num_rows($query);
    if($numrow !=0){
        while($row = mysql_fetch_assoc($query)){
            $dbusername = $row['username'];
            $dbpassword = $row['password'];         
        }
        if($username == $dbusername && $password == $dbpassword ){
            $_SESSION['username'] == $dbusername;
            header("Location: userhome.php");
        }
        else{
            echo "Incorect password";
        }
    }
    else{
        die("This user dosent exists");
    }
}
else{
    $reg = die("Please enter username and password");
}
}

?>

在我自己的意见有问题的地方在SESSION或查询，但我已经通过了许多不同的资源，它不应该是一个问题....我可能错过了一些非常基本的东西或一些蹩脚的错误:D

很简单。您使用if(isset($_POST['Edit Account']))，但您的提交按钮没有名称:<input type="submit" value="Edit Account" class="button">

将输入改为:

<input type="submit" value="Edit Account" name="edit_account" class="button">

，将if(isset($_POST['Edit Account']))改为if(isset($_POST['edit_account']))

$edit在哪里?

WHERE username='$edit'

也许你必须修改会话的$dbusername

Hovewer:

-不要使用mysql，使用mysqli或pdo

-您可以在一行中保护所有变量

foreach($_POST as $key => $var) $_POST[$key] = mysqli_real_escape_string($var);

-添加对输入变量的验证控制，以及登录会话检查