php - 使用 PHP、Ajax 和 Jquery 更改密码

Question

我想我需要一个有更好眼光的人来帮助我发现我的错误。我正在尝试更改当前密码而不刷新页面。我收到错误消息“错误，请重试!”。这是我到目前为止的代码。

HTML

<form name="resetform" action="changepass.php" id="resetform" class="passform" method="post" role="form">
  <h3>Change Your Password</h3>
  <br />
  <input type="hidden" name="username" value="<?php echo $sname;?>" ></input>
  <label>Enter Old Password</label>
  <input type="password" class="form-control" name="old_password" id="old_password">
  <label>Enter New Password</label>
  <input type="password" class="form-control" name="new_password" id="new_password">
  <label>Confirm New Password</label>
  <input type="password" class="form-control"  name="con_newpassword"  id="con_newpassword" />
 <br>
 <input type="submit" class="btn btn-warning" name="password_change" id="submit_btn" value="Change Password" />
</form>

PHP

<?php
include_once 'database-config.php';

if (isset($_POST['password_change'])) {

    $username = strip_tags($_POST['sname']);
    $password = strip_tags($_POST['old_password']);
    $newpassword = strip_tags($_POST['new_password']);
    $confirmnewpassword = strip_tags($_POST['con_newpassword']);

// match username with the username in the database
    $sql = "SELECT * FROM users WHERE username='$sname'";

    $query = $dbh->prepare($sql);

    $query->execute();

    $row = $query->fetchAll();

    $hash = $row[0]["password"];

    //$hash = $results[0]["password"];

        if ($password == $hash){

            if($newpassword == $confirmnewpassword) {

            $sql = "UPDATE users SET password = '$newpassword' WHERE    username  = '$username'";

            $query = $dbh->prepare($sql);

            $query->execute();

            echo "Password Changed Successfully!";

        } else echo "Passwords do not match!";
        }
    } else echo "Please type your current password accurately!";
}
?>

Jquery

<script type="text/javascript">
$(document).ready(function() {
        var frm = $('#resetform');
        frm.submit(function(e){
            $.ajax({
                type: frm.attr('method'),
                url: frm.attr('action'),
                data: frm.serialize(),
                success: function(data){
                    $('#success').html("<div id='message'></div>");
                    $('#message').html("<h2>Password changed successfully!</h2>")
                    .delay(3000).fadeOut(3000);
                },
                error: function(data) {
                    $('#error').html("<div id='errorMessage'></div>");
                    $('#errorMessage').html("<h3>Error, please try again!</h3>")
                    .delay(2000).fadeOut(2000);
                }

            });

            e.preventDefault();
        });
    });
</script>

如果有任何更正，我将不胜感激:-)

Answer 1

您的代码存在多个问题，例如:

• 请参阅此处的声明，

  if (isset($_POST['password_change'])) { ...
  

  这里，$_POST['password_change']没有设置，因为jQuery的serialize()不包括编码按钮或提交输入，所以你必须附加名称以及结果的提交按钮的值，如下所示:

  var formData = frm.serialize();
  formData += '&' + $('#submit_btn').attr('name') + '=' + $('#submit_btn').attr('value');
  

• 由于此语句，变量 $username 未设置，

  $username = strip_tags($_POST['sname']);
  

  应该是

  $username = strip_tags($_POST['username']);
  

• 没有必要在另一个元素内创建一个元素来显示成功/错误消息，而且无论查询结果如何，该消息都是相同的。另外，您没有使用服务器的响应数据。在 AJAX 请求中查看以下回调函数，

  success: function(data){
      $('#success').html("<div id='message'></div>");
      $('#message').html("<h2>Password changed successfully!</h2>")
      .delay(3000).fadeOut(3000);
  },
  error: function(data) {
      $('#error').html("<div id='errorMessage'></div>");
      $('#errorMessage').html("<h3>Error, please try again!</h3>")
      .delay(2000).fadeOut(2000);
  }
  

  相反，创建一个 div 元素，如下所示:

  <div id="message"></div>
  

  并在回调函数中显示成功/错误消息，如下所示:

  success: function(data){
      $('#message').html(data).delay(3000).fadeOut(3000);
  },
  error: function(jqXHR, textStatus, errorThrown) {
      $('#message').html(textStatus).delay(2000).fadeOut(2000);
  }
  

• 您的 PHP 代码中存在一些语法错误，包括解析错误:语法错误，...中出现意外的“}”。

• 始终准备、绑定(bind)和执行您的查询，以防止任何类型的 SQL 注入(inject)。 And this is how you can prevent SQL injection in PHP .

• 切勿将密码存储为纯可读文本，始终执行 salted password hashing在将原始密码插入到表中之前。

所以你的代码应该是这样的:

HTML:

<form name="resetform" action="changepass.php" id="resetform" class="passform" method="post" role="form">
    <h3>Change Your Password</h3>
    <br />
    <input type="hidden" name="username" value="<?php echo $sname; ?>" ></input>
    <label>Enter Old Password</label>
    <input type="password" class="form-control" name="old_password" id="old_password">
    <label>Enter New Password</label>
    <input type="password" class="form-control" name="new_password" id="new_password">
    <label>Confirm New Password</label>
    <input type="password" class="form-control"  name="con_newpassword"  id="con_newpassword" />
    <br>
    <input type="submit" class="btn btn-warning" name="password_change" id="submit_btn" value="Change Password" />
</form>

<!--display success/error message-->
<div id="message"></div>

jQuery:

$(document).ready(function() {
    var frm = $('#resetform');
    frm.submit(function(e){
        e.preventDefault();

        var formData = frm.serialize();
        formData += '&' + $('#submit_btn').attr('name') + '=' + $('#submit_btn').attr('value');
        $.ajax({
            type: frm.attr('method'),
            url: frm.attr('action'),
            data: formData,
            success: function(data){
                $('#message').html(data).delay(3000).fadeOut(3000);
            },
            error: function(jqXHR, textStatus, errorThrown) {
                $('#message').html(textStatus).delay(2000).fadeOut(2000);
            }

        });
    });
});

PHP:

<?php

    include_once 'database-config.php';

    if (isset($_POST['password_change'])) {

        $username = strip_tags($_POST['username']);
        $password = strip_tags($_POST['old_password']);
        $newpassword = strip_tags($_POST['new_password']);
        $confirmnewpassword = strip_tags($_POST['con_newpassword']);

        // match username with the username in the database
        $sql = "SELECT * FROM `users` WHERE `username` = ? LIMIT 1";

        $query = $dbh->prepare($sql);
        $query->bindParam(1, $username, PDO::PARAM_STR);

        if($query->execute() && $query->rowCount()){
            $hash = $query->fetch();
            if ($password == $hash['password']){
                if($newpassword == $confirmnewpassword) {
                    $sql = "UPDATE `users` SET `password` = ? WHERE `username` = ?";

                    $query = $dbh->prepare($sql);
                    $query->bindParam(1, $newpassword, PDO::PARAM_STR);
                    $query->bindParam(2, $username, PDO::PARAM_STR);
                    if($query->execute()){
                        echo "Password Changed Successfully!";
                    }else{
                        echo "Password could not be updated";
                    }
                } else {
                    echo "Passwords do not match!";
                }
            }else{
                echo "Please type your current password accurately!";
            }
        }else{
            echo "Incorrect username";
        }
    }

?>