Question

我正在制作的网站的一部分为用户提供了更改密码的选项。我已配置此代码，它工作正常。

我现在想要将一个表单添加到同一页面，这样就可以让用户选择更改其USERNAME。（单独地，所以他们更改用户名但不更改密码）。

我是否需要重复完全相同的以下代码并将其粘贴到其下方，显然需要使用特定字词作为用户名而不是密码，或者我是否可以添加到现有代码这样它的单一处理代码？我该怎么做呢？例如，第一行：

if（$ submit =='Change Password'）{，

在我脑子里用英语，我想要代码说 IF '更改密码'或'更改用户名'按钮已提交...以下代码等。

//处理代码

<?php
require_once('checklog.php');
require_once('functions.php');
// Grab the form data
$submit         = trim($_POST['submit']);
$password       = trim($_POST['password']);
$newpassword    = trim($_POST['newpassword']);
$repeatpassword = trim($_POST['repeatpassword']);
// Create some variables to hold output data
$message        = '';
// Start to use PHP session
session_start();
if ($submit == 'Change Password') {
    include_once('connect.php');
    mysqli_select_db($db_server, $db_database) or die("Couldn't find db");
    //clean the input now that we have a db connection                      
    $password       = clean_string($db_server, $password);
    $newpassword    = clean_string($db_server, $newpassword);
    $username       = $_SESSION['username'];
    $repeatpassword = clean_string($db_server, $repeatpassword);
    if ($password && $newpassword) {
        if ($repeatpassword == $newpassword) {
            if (strlen($newpassword) > 25 || strlen($newpassword) < 6) {
                $message = "Password must be 6-25 characters long";
            } else {
                // check whether username exists
                $password = salt($password);
                $query    = "SELECT username FROM users WHERE username='$username' and password='$password'";
                $result   = mysqli_query($db_server, $query);
                //if theres a result change password to new password
                if ($row = mysqli_fetch_array($result)) {
                    $newpassword    = salt($newpassword);
                    $repeatpassword = salt($repeatpassword);
                    //update password
                    $query          = "UPDATE users SET password='$newpassword' WHERE username='$username'";
                    mysqli_query($db_server, $query) or die("Update failed. " . mysqli_error($db_server));
                    $message = "<strong>Password change successful!</strong>";
                } else {
                    $message = "User account not found.";
                }
                mysqli_free_result($result);
            }
        } else {
            $message = "Password must match";
        }
    } else {
        $message = "Please enter all fields";
    }
    include_once('db_close.php');
}


include_once('header_logged.php');

?>

//更改密码或用户名的表单

!--change password form --> 
To change your Password:

<form method="post" action="changepassword.php">
Current password: <input type='password' name='password'>
<br>
<br>
New Password: <input type='password' name='newpassword'>
<br>
<br>
Repeat New Password: <input type='password' name='repeatpassword'>

<input type='submit' name='submit' value='Change Password'>
<input type='reset' name='reset' value='Reset'>
</form>
<p><strong><?php
echo $message;
?></strong></p>


<!--change username form --> 
To change your Username:
<form method="post" action="changepassword.php">
Current Username: <input type='username' name='username'>
<br>
<br>
New Username: <input type='username' name='newusername'>
<br>
<br>
Repeat New Username: <input type='username' name='repeatusername'>

<input type='submit' name='change' value='Change Username'>
<input type='reset' name='reset' value='Reset'>
</form>

感谢您的帮助，非常感谢。

Answer 1

使用它：

if(isset($_POST['submit']) || isset($_POST['change'])){
//write your code here
}

如何编写代码以允许用户更改密码或更改用户名？

1 个答案: