Question

我试图创建一个使用jQuery AJAX和PHP的联系表单。我的HTML如下

<form action="#" id="contactForm" method="post" name="contactForm">
<fieldset>
    <div class="col-sm-12">
        <input id="contactName" name="contactName" placeholder="Your Name*" type="text" value="" required>
    </div>

    <!-- Name Field [ END ] -->

    <div class="col-sm-12">
        <input id="contactEmail" name="contactEmail" placeholder="Your Email*" type="email" value="" required>
    </div>

    <!-- Email Field [ END ] -->

    <div class="col-xs-12">
        <textarea cols="5" id="contactMessage" name="contactMessage" placeholder="Your Message....*" required></textarea>
    </div>

    <!-- Message Field [ END ] -->

    <div class="col-xs-12">
        <button class="submit">SEND</button>
    </div>

    <!-- Submit Button [ END ] -->

    <div class="error col-xs-12">
        <h3>Sorry! Your message was not sent.</h3>
    </div>

    <!-- Error Message [ END ] -->

    <div class="success col-xs-12">
        <h3>Success! Your message was sent.</h3>
    </div>

    <!-- Seccess Message [ END ] -->

</fieldset>

虽然jQuery代码如下

$('form#contactForm').submit(function () {

    var url = $('#contactForm').attr('action'),
        type = $('#contactForm').attr('method'),

        contactName = $('#contactForm #name').val(),
        contactEmail = $('#contactForm #email').val(),
        contactMessage = $('#contactForm #message').val(),
        data = {
            name: contactName,
            email: contactEmail,
            message: contactMessage
        };

    $.ajax({
        type: type,
        url: url,
        data: data,
        success: function () {
            $('.success').fadeIn();
            $('.error').fadeOut();
        },
        error: function () {
            $('.error').fadeIn();
            $('.success').fadeOut();
        }
    });
    return false;
});

什么可能的PHP代码使这个表单工作。我正在使用以下代码

 <?php

    if ( isset( $_POST['name'], $_POST['email'], $_POST['message'] ) ) {

        $to = 'mail@example.com'; //my email address here
        $subject = 'Message From :' . $_POST['name'];
        $message = $_POST['message'];
        $headers = 'From :' . $_POST['email'].'\r\n';

        mail($to,$subject,$message,$headers);

    }
?>

但它不起作用。成功消息显示但未收到邮件。我不太熟悉PHP。

请帮助......

Answer 1

您的成功消息显示因为php-page被成功调用，这并不表示邮件已成功发送。您应该通过返回邮件功能响应来检查：

$response = mail($to,$subject,$message,$headers);
echo $response;

这将返回true或false。此外，您还必须检查邮件是否正确接收，最终可能是垃圾邮件，还是由于各种原因被邮件服务器拒绝。如果是这种情况，您可以对邮件头进行一些添加。关于如何通过

获取邮件的Here are some pointers

Answer 2

尝试添加

method: "POST",

在$.ajax来电中，默认情况下，$ .ajax使用GET请求

Answer 3

这对你有用

$('form#contactForm').submit(function () {

    var url = $('#contactForm').attr('action');
        var type = $('#contactForm').attr('method');

        var contactName = $('#contactName').val(); 
        var contactEmail = $('#contactEmail').val(); 
         var contactMessage = $('#contactMessage').val();
        data = {
            name: contactName,
            email: contactEmail,
            message: contactMessage
        };console.log(data);

    $.ajax({
        type: type,
        url: url,
        data: data,
        success: function () {
            $('.success').fadeIn();
            $('.error').fadeOut();
        },
        error: function () {
            $('.error').fadeIn();
            $('.success').fadeOut();
        }
    });
    return false;
});

您的PHP代码变为

<?php
if ( isset( $_POST['name'], $_POST['email'], $_POST['message'] ) ) {

    $to = $_POST['email']; //my email address here
    $subject = 'Message From :' . $_POST['name'];
    $message = $_POST['message'];
    $headers = 'From :' . $_POST['email'].'\r\n';

    mail($to,$subject,$message,$headers);

}
        echo 'sucess';
?>

也在这里改变

<form action="===GIVE YOUR FILE NAME CONTAINING PHP mail() CONTAINS====" id="contactForm" method="post" name="contactForm">

您应确认您的服务器已启用mail()功能

试试这个让我知道..

AJAX联系表单的{PHP代码}

3 个答案: