Question

这是我的javascirpt代码

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(6000).fadeOut();
    });
});

我没有得到post函数无法获取的字段值

Answer 1

尝试将方法设置为＆＃34; POST＆＃34;和$ .ajax中的数据，如下所示：

$.ajax({
    method: "POST",
    data: form.serialize(),
    etc...

此外，查看PHP文件会有所帮助。

Answer 2

<div class="contact-form wow fadeIn" data-wow-duration="1000ms" data-wow-delay="600ms">
      <div class="row">
        <div class="col-sm-6">
          <form id="main-contact-form" name="contact-form" method="post" action="sendemail.php">
            <div class="row  wow fadeInUp" data-wow-duration="1000ms" data-wow-delay="300ms">
              <div class="col-sm-6">
                <div class="form-group">
                  <input type="text" id="name" name="name" class="form-control" placeholder="Name" required="required">
                </div>
              </div>
              <div class="col-sm-6">
                <div class="form-group">
                  <input type="email" id="email" name="email" class="form-control" placeholder="Email Address" required="required">
                </div>
              </div>
            </div>
            <div class="form-group">
              <input type="text" id="subject" name="subject" class="form-control" placeholder="Subject" required="required">
            </div>
            <div class="form-group">
              <textarea name="message" id="message" class="form-control" rows="4" placeholder="Enter your message" required="required"></textarea>
            </div>                        
            <div class="form-group">
              <button type="submit" class="btn-submit">Send Now</button>
            </div>
          </form>   
        </div>

这是我的HTML代码

Answer 3

您似乎只是发送一个空的ajax请求。您需要方法和表单数据才能发送，例如：

$.ajax({
    url: $(this).attr('action'),
    type: 'post',
    dataType: 'json', // <-- not sure if about this
    data: form.serialize(),
beforeSend: function(){
    ...

另见Submit a form using jQuery

PHP联系表单，帖子功能不起作用

3 个答案: