Question

所以我正在制作注册表格，我有一个ajax脚本来验证和提交数据。这是代码：

$('document').ready(function() {
    /* validation */
    $("#register-form").validate({
        rules: {
            username: {
                required: true,
                rangelength: [4, 12]
            },
            email: {
                required: true,
                email: true
            },
            password: {
                required: true,
                rangelength: [4, 20]

            },
            password_repeat: {
                required: true,
                equalTo: "#password"
            },
        },
        messages: {
            username: {
                required: "*Моля въведете потребителско име",
                rangelength: "*Между 4 и 12 знака"

            },
            email: {
                required: "*Моля въведете имейл"

            },
            password: {
                required: "*Моля въведете парола",
                rangelength: "*Между 8 и 20 знака"

            },
            password_repeat: {

                required: "*Моля въведете паролата отново",
                equalTo: "*Паролите не съвпадат"

            },
        },

        submitHandler: submitForm
    });

    /* Handling login functionality */
    function submitForm() {
        var data = $("#register-form").serialize();
        $.ajax({
            type: 'POST',
            url: 'auth_logic/register-process.php',
            data: data,
            beforeSend: function() {
                $("#error").fadeOut();
                $("#register").html('<img src="images/loading.gif" width="31px" height="31px" /> &nbsp; Моля изчакайте ...');
            },
            success: function(response) {
                if (response == "ok") {
                    $("#register").html('<img src="images/loading.gif" width="31px" height="31px" /> &nbsp; Моля изчакайте ...');
                    $("#error").html('<div class="alert alert-danger"> &nbsp; ' + response + '</div>');
                    //setTimeout(' window.location.href = "/auth/login"; ',100);
                } else if (response == "Потребителското име е заето!") {
                    $("#error").fadeIn(100, function() {
                        $("#error").html('<div class="alert alert-danger"> &nbsp; ' + response + '</div>');
                        $("#register").html('&nbsp; Регистрация');
                    });
                } else if (response == "Този имейл е регистриран вече!") {
                    $("#error").fadeIn(100, function() {
                        $("#error").html('<div class="alert alert-danger">  &nbsp; ' + response + '</div>');
                        $("#register").html(' &nbsp; Регистрация');
                    });
                }
            }
        });

        return false;
    }
});

PHP部分工作正常但是当我想更新＆＃34; #error＆＃34; div我在表单之前没有工作，但它更新了提交按钮。

以下是表格：

<div class="container py-5">
<div class="row">
    <div class="col-md-12">
        <div class="row">
            <div class="col-md-6 mx-auto">
                <div class="card border-secondary">
                    <div class="card-header">
                        <center><h3 class="mb-0 my-2">Нов потребител</h3></center>
                    </div>
                    <div class="card-body">
                        <div id="error"></div>
                        <form class="form-group" method="POST" id="register-form">                              
                            <div class="form-group">
                                <label for="username">Потребител:</label>
                                <input type="text" class="form-control" name="username" id="username" placeholder="Потребителско име" >
                            </div>
                            <div class="form-group">
                                <label for="email">Имейл:</label>
                                <input type="email" class="form-control" name="email" id="email" placeholder="Имейл адрес" >
                            </div>
                            <div class="form-group">
                                <label for="password">Парола:</label>
                                <input type="password" class="form-control" name="password" id="password" placeholder="Парола">
                            </div>
                            <div class="form-group">
                                <label for="password_again">Парола отново:</label>
                                <input type="password" class="form-control" name="password_repeat" id="password_repeat" placeholder="Повторете паролата" >
                            </div>
                            <div class="form-group">
                                <button type="submit" name="register" id="register" class="btn btn-block btn-success btn-lg float-right">Регистрация</button>
                            </div>
                        </form>
                    </div>
                    <div class="card-footer">
                        <a href="login">Вход</a>
                    </div>                        
                </div>
            </div>
        </div>
        <!--/row-->

    </div>
    <!--/col-->
</div>
<!--/row-->

我清除缓存刷新了400次，重命名文件。我正在使用XAMPP

Answer 1

由于beforeSend:隐藏#error .fadeOut();

在success:上，你替换了#error中的内容，但它仍然不可见。您需要添加$('#error').show();或包含在内的内容：

$("#error").fadeIn(100, function() {
}

AJAX寄存器不会更新div

1 个答案: