Question

当我点击按钮加载js文件时，没有任何反应，我已经单独测试了php文件而没有ajax只是纯PHP而且工作正常！所以我认为这里的代码存在问题：

$(document).ready(function(){
    $('#changePasswordNeededButton').click(function(){
        $.post("http://example.com/change_password.php",
        {
            currentPassword : $('#currentPassword').val(),
            newPassword : $('#newPassword').val(),
            repeatNewPassword : $('#repeatNewPassword').val()
        },
            function(data){
                if (data.result == 1) {
                    location.href = data.location;
                }
            }, "json" // get content which has been returned in json format
        );
    });
});

HTML for this：

<div class='changePassword'>
<div id='changePasswordInner'>
    <form>
        <input id='currentPassword' type='password' name='currentPassword' placeholder='Nuværende adgangskode'>
        <input id='newPassword' type='password' name='newPassword' placeholder='Ny adgangskode'>
        <input id='repeatNewPassword' type='password' name='repeatNewPassword' placeholder='Gentag ny adgangskode'>
        <input id='changePasswordNeededButton' type='button' name='changePasswordNeededButton' value='Skift adgangskode'>
    </form>
    <div id'errorChangePassword'></div>
</div>
</div>

<script src="http://code.jquery.com/jquery-2.1.4.min.js"></script>
<script src="../script/change_password.js"></script>

我的php：

if (isset($_POST['currentPassword']) && isset($_POST['newPassword']) && isset($_POST['repeatNewPassword']) {

    $currentPassword = $_POST['currentPassword'];
    $newPassword = $_POST['newPassword'];
    $repeatNewPassword = $_POST['repeatNewPassword'];

    $query = "SELECT * FROM user WHERE password = '$currentPassword'";
    $result = $db->query($query);
    while ($row = $result->fetch_assoc()) {
        $password = $row['password'];
    }

    if ($password == $currentPassword) {
        $update = "UPDATE user SET password = '$newPassword' WHERE password = '$currentPassword'";
        $db->query($update);
        $result = array(
            'result' => 1,
            'location' => 'index.php',
        );
    }

    // return the result
        echo json_encode($result);
        die();
}

我希望有人能在这里看到错误吗？

谢谢..

Answer 1

看一下jQuery的$ .post文档。我实际上会争论使用$ .ajax而不是因为它带来了额外的功能;主要是为成功和失败添加回调的能力。

来自jQuery的AJAX文档：

..

Answer 2

检查此行

'location' => 'index.php',

您的代码不适用于额外的, at the end of that line。由于您的脚本需要json并且收到格式错误的json字符串。

        <?php

    if (isset($_POST['currentPassword']) && isset($_POST['newPassword']) && isset($_POST['repeatNewPassword'])) {



     $result = array(
                'result' => 1,
                'location' => 'index.php'
            );

        echo json_encode($result);

    exit(); 
    };

    ?>

     <form>
            <input id='currentPassword' type='password' name='currentPassword' placeholder='Nuværende adgangskode'>
            <input id='newPassword' type='password' name='newPassword' placeholder='Ny adgangskode'>
            <input id='repeatNewPassword' type='password' name='repeatNewPassword' placeholder='Gentag ny adgangskode'>
            <input id='changePasswordNeededButton' type='submit' name='changePasswordNeededButton' value='Skift adgangskode'>
    </form>

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

    <script type="text/javascript">

    $(document).ready(function(){
        $('#changePasswordNeededButton').click(function(){
            $.post("",
            {
                currentPassword : $('#currentPassword').val(),
                newPassword : $('#newPassword').val(),
                repeatNewPassword : $('#repeatNewPassword').val()
            },
                function(data){


                    if (data.result == 1) {
                    location.href=data.location;
                    }


                },"json"
            );
            return false;
        });
    });
    </script>

ajax电话无法查看错误：S

2 个答案: