PHP的联系方式发送电子邮件，没有任何信息

Question

我正在从网站的联系表中收到电子邮件，但未收到任何内容。

这是我的代码：

的PHP

<?php

$name       = @trim(stripslashes($_POST["name"])); 
$from       = @trim(stripslashes($_POST["email"])); 
$subject    = @trim(stripslashes($_POST["subject"])); 
$message    = @trim(stripslashes($_POST["message"])); 

// recipient address
$to = "admin@deptfordtechservices.com";

// subject of email
$re = "New Inquiry";

// message creation
$contact = "Name:".$name."\nEmail:".$from."\nSubject:".$subject."\r\n";

$txt = "Message:".$message."\r\n";


$fmsg = $contact."\r\n".$txt;

$msg = wordwrap($fmsg, 70);

// send email
mail($to,$re,$msg);

die;

的HTML

<form id="main-contact-form" name="contact-form" method="post" action="sendemail.php">
    <div class="form-group">
        <input type="text" name="name" id="name" class="form-control" placeholder="Name" required>
    </div>
    <div class="form-group">
        <input type="email" name="email" id="email" class="form-control" placeholder="Email" required>
    </div>
    <div class="form-group">
        <input type="text" name="subject" id="subject" class="form-control" placeholder="Subject" required>
    </div>
    <div class="form-group">
        <textarea name="message" id="message" class="form-control" rows="8" placeholder="Message" required></textarea>
    </div>
    <button type="submit" class="btn btn-primary">Send Message</button>
</form>

Java脚本

// Contact form
var form = $('#main-contact-form');
form.submit(function(event){
    event.preventDefault();
    var form_status = $('<div class="form_status"></div>');
    $.ajax({
        url: $(this).attr('action'),
        beforeSend: function(){
            form.prepend( form_status.html('<p><i class="fa fa-spinner fa-spin"></i> Email is sending...</p>').fadeIn() );
        }
    }).done(function(data){
        form_status.html('<p class="text-success">Thank you for contact us. As early as possible  we will contact you</p>').delay(3000).fadeOut();
    });
});

网站：deptfordtechservices.com

提前致谢，

大卫

Answer 1

Javascript未发送表单输入。 它需要一个data:选项来提供参数，并且需要使用type: 'POST'因此它发送POST而不是GET 。

// Contact form
var form = $('#main-contact-form');
form.submit(function(event){
    event.preventDefault();
    var form_status = $('<div class="form_status"></div>');
    $.ajax({
        url: $(this).attr('action'),
        type: 'POST',
        data: $(this).serialize(),
        beforeSend: function(){
            form.prepend( form_status.html('<p><i class="fa fa-spinner fa-spin"></i> Email is sending...</p>').fadeIn() );
        }
    }).done(function(data){
        form_status.html('<p class="text-success">Thank you for contact us. As early as possible  we will contact you</p>').delay(3000).fadeOut();
    });
});