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Javascript 发送消息后弹出窗口不工作_Javascript_Jquery_Html_Css - Fatal编程技术网

Javascript 发送消息后弹出窗口不工作

Javascript 发送消息后弹出窗口不工作,javascript,jquery,html,css,Javascript,Jquery,Html,Css,在我的“联系我们”页面中,当我发送消息时,弹出窗口将显示并消失,但什么也没有发生。请帮忙 “联系我们”页面和弹出模式 <!-- The Contact Us Page --> <div class="jumbotron jumbotron-sm"> <div class="container"> <div class="row"&g

在我的“联系我们”页面中,当我发送消息时,弹出窗口将显示并消失,但什么也没有发生。请帮忙

“联系我们”页面和弹出模式

              <!-- The Contact Us Page -->
              <div class="jumbotron jumbotron-sm">
                <div class="container">
                  <div class="row">
                    <div class="col-sm-12 col-lg-12 header"></div>
                  </div>
                </div>
              </div>
              <div class="container card">
                <div class="row">
                  <div class="col-sm-12 col-lg-12">
                    <h1 class="h1">Contact us <span class="fa fa-envelope"></span> <!--<small class="col-xs-12" style="padding-left: 0">We will get back to You</small>--></h1>
                  </div>
                </div>
                <div class="row">
                  <div class="col-sm-12 col-md-12">
                    <div class="cardStyle">
                      <form name="contact" id="form" data-toggle="validator">
                        <div class="row">
                          <div class="col-md-6">
                            <div class="form-group">
                              <label for="name">
                                Name</label>
                              <input type="text" class="form-control" id="name" placeholder="Enter name" required="required" />
                            </div>
                            <div class="form-group">
                              <label for="email">
                                Email Address</label>
                              <div class="input-group">
                                <span class="input-group-addon"><span class="glyphicon glyphicon-envelope"></span>
                                </span>
                                <input type="email" class="form-control" id="email" placeholder="Enter email" required="required" />
                              </div>
                            </div>
                            <div class="form-group">
                              <label for="subject">
                                Subject</label>
                              <select id="subject" name="subject" class="form-control" required="required">
                                <option value="none" selected="" disabled>Choose One:</option>
                                <option value="general">General Customer Service</option>
                                <option value="collaborate">Collaborate with Us</option>
                                <option value="bug">Found a Bug/Issue</option>
                                <option value="other">Any other Queries</option>
                              </select>
                            </div>
                          </div>
                          <div class="col-md-6">
                            <div class="form-group">
                              <label for="name">
                                Message</label>
                              <textarea name="message" id="message" class="form-control" rows="9" cols="25" required="required" placeholder="Message"></textarea>
                            </div>
                          </div>
                          <div class="col-md-12">
                            <button type="submit" class="btn btn-success pull-right" id="btnContactUs">Send Message</button>
                          </div>
                        </div>
                      </form>
                    </div>
                  </div>
                </div>
              </div>

由于您仅在未调用
preventDefault()
时才尝试显示模式,因此表单的默认操作将发生,即提交表单。这会导致页面加载,并且由于您尚未在表单上指定
操作
属性,因此它将提交到同一页面。换句话说,页面只是重新加载


如果不想导致页面重新加载,则必须始终调用
preventDefault()
,然后使用AJAX提交数据。

由于您仅在未调用
preventDefault()
时才尝试显示模式,因此表单的默认操作将发生,即提交表单。这会导致页面加载,并且由于您尚未在表单上指定
操作
属性,因此它将提交到同一页面。换句话说,页面只是重新加载

如果不想导致页面重新加载,则必须始终调用
preventDefault()
,然后使用AJAX提交数据

                                <!-- Form submitted Thank You Modal -->
                                    <div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel">
                                    <div class="modal-dialog" role="document">
                                    <div class="modal-content">
                                    <div class="modal-header">
                                        <button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">&times;</span></button>
                                        <h4 class="modal-title text-center" id="myModalLabel">Contact</h4>
                                    </div>
                                    <div class="modal-body">
                                        <h3 class="h3 text-center">Thank you for your feedback! We will get back to you.</h3>
                                    </div>
                                    <div class="modal-footer">
                                        <button type="button" class="btn btn-success" data-dismiss="modal">Close</button>
                                    </div>
                                    </div>
                                    </div>
                                    </div>
    $(document).ready(function() {
      $('#form').validator().on('submit', function(e) {
        if (e.isDefaultPrevented()) {
          // handle the invalid form...
          $('#myModal').modal('hide');
        } else {
          // everything looks good!
          $('#myModal').modal('show');
        }
      })

    });