PHP表单AJAX“的成功和放大器;失败“的模态窗口消息 [英] Php form ajax “success & fail” message on modal window

问题描述

我不能在同一个模态窗口中的错误信息或成功。 我已经安装了字preSS和code我使用的是如下：

I can not make the error message or success in the same modal window. I have installed wordpress and the code I'm using is as follows:

表格

<form id="formcotizador" method="post" action="<?php echo get_template_directory_uri(); ?>/cotizador/proccess.php">                             

    <div class="small-12 medium-4 large-4 columns">
        <input id="nombre" type="text" name="nombre" placeholder="Nombre y Apellido" required>
    </div>

    <div class="small-12 medium-4 large-4 columns">
        <select name="provincia" id="provincia">
            <option value="default">Seleccione una Provincia</option>
            <option value="Buenos Aires">Buenos Aires</option>
            <!-- ... -->
        </select>
    </div>

    <div class="small-12 medium-4 large-4 columns">
        <input type="text" id="localidad" name="localidad" placeholder="Localidad" required>
    </div>

    <div class="small-12 medium-4 large-4 columns">
        <input type="email" id="email" name="email" placeholder="Tu email" required>
    </div>

    <div class="small-12 medium-4 large-4 columns">
        <input type="submit" value="Enviar consulta" class="button tiny radius">
    </div>
    <!-- error, success div -->
    <div class="small-12 medium-4 large-4 columns msjj"></div>

</form>

PHP （proccess.php）

PHP (proccess.php)

<?php 

    if ($_SERVER['REQUEST_METHOD'] == 'POST') {
        $name = strip_tags(trim($_POST['nombre']));
        $name = str_replace(array("\r","\n"),array(" "," "),$name);
        $provincia = $_POST['provincia'];
        $localidad = strip_tags(trim($_POST['localidad']));
        $email = filter_var(trim($_POST["email"]), FILTER_SANITIZE_EMAIL);
        $telefono = strip_tags(trim($_POST['telefono']));

        $recipient = "[e-mail]";

        // Set the email subject.
        $subject = "New contact from $name";

        // Build the email content.
        $email_content = "Name: $name\n";
        $email_content .= "Email: $email\n\n";
        $email_content .= "Message:\n$message\n";

        // Build the email headers.
        $email_headers = "From: $name <$email>";

        if (mail($recipient, $subject, $email_content, $email_headers)) {
            http_response_code(200);
            echo "Gracias por enviar el msj";
        } else {
            http_response_code(500);
            echo "Hubo un error al enviar el msj";
        }


    } else {
        http_response_code(403);
        echo "hubo un problema al enviar, pruebe de nuevo";
    }
?>

JS /阿贾克斯

$(function() {

    var form = $('#formcotizador');
    var formMessages = $('.msjj');

    $(form).submit(function(e) {
        e.preventDefault();

        var formData, = $(form).serialize();

        $.ajax({
            type: 'POST',
            url: $(form).attr('action'),
            data: formData,         
        })
        .done(function(response) {          
                $(formMessages).removeClass('error');
                $(formMessages).addClass('success');

                $(formMessages).text(response);

                $('#nombre').val('');
                $('#provincia').val('default');
                $('#localidad').val('');
                $('#email').val('');
        })
        .fail(function(data) {

                $(formMessages).removeClass('success');
                $(formMessages).addClass('error');

                if (data.responseText !== '') {
                    $(formMessages).text(data.responseText);
                } else {
                    $(formMessages).text('Este es el error');
                }; 
        });
    });
});

当我送，我重定向到形式'行动'的属性页，从proccess.php

When I send, I redirected to the page of the form 'action' attribute, with the message from the proccess.php

您可以测试自己这里。

这是在code表示修改：如何创建一个AJAX触点形式。 我认为这个问题是该职位没有解释不是一个模态窗口。

This is the code that modifies: How to Create an AJAX Contact Form. I think the problem is that the post is not explained not for a modal window.

推荐答案

您有您的网页上几个JS错误。但与此code特别

You have a few JS errors on your page. But with this code in particular

var formData, = $(form).serialize();

应

var formData = $(form).serialize();

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