php - 使用PHP，Ajax和Jquery更改密码

我认为我需要一个眼睛更好的人来帮助我发现我的错误。我试图更改当前密码而不刷新页面。我收到错误消息“错误，请重试！”。到目前为止，这是我的代码。

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>

的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的

<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>

我将不胜感激任何纠正:-)

最佳答案

您的代码有几个问题，例如：


在这里查看此声明，

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

在这里，未设置$_POST['password_change']，因为jQuery的serialize()不包括编码按钮或提交输入，因此您必须在结果中附加提交按钮的名称和值，如下所示：

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

由于该语句，未设置变量$username，

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

它应该是，

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

在另一个元素内创建一个元素来显示成功/错误消息是没有意义的，并且无论查询结果如何，消息都是相同的。另外，您没有利用服务器的响应data。请在您的AJAX请求中查看以下这些回调函数，

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，制作一个div元素，如下所示：

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

并在您的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);
}

在您的PHP代码中，几乎没有由Parse error: syntax error, unexpected '}' in ...组成的语法错误。
始终准备，绑定和执行查询，以防止进行任何形式的SQL注入。 And this is how you can prevent SQL injection in PHP。
切勿将密码存储为易读的文本，在将原始密码插入表中之前，请始终对原始密码执行salted password hashing。


因此，您的代码应如下所示：

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的：

$(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";
        }
    }

?>

关于php - 使用PHP，Ajax和Jquery更改密码，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/37168637/