每次选择或单击下拉项时如何使用 AJAX Post 方法执行发送请求

Question

每次选择或单击下拉项时如何使用 AJAX Post 方法执行发送请求，这是我的 HTML 代码

<div class="dropdown-menu dropdown-large" aria-labelledby="navbarDropdown">
                        <div class="row">
                            <div class="col-sm-3">
                                <a class="dropdown-item"><b>Title 1</b></a>
                                <a class="dropdown-item" href="#">dp1</a>
                                <a class="dropdown-item" href="#">dp2</a>
                                <a class="dropdown-item" href="#">dp3/a>
                                <a class="dropdown-item" href="#">dp4</a>
                                <a class="dropdown-item" href="#">dp5</a>
                            </div>                           
                        </div>

                    </div>

Answer 1

两种可能：

1. 使用事件监听器
2. 将 onClick 添加到您的下拉元素中

Answer 2

html

<div class="dropdown-menu dropdown-large" aria-labelledby="navbarDropdown">
  <div class="row">
    <div class="col-sm-3">
     <a class="dropdown-item"><b>Title 1</b></a>
     <a class="dropdown-item" name="tryme" href="#">dp1</a>
     <a class="dropdown-item" name="tryme" href="#">dp2</a>
     <a class="dropdown-item" name="tryme" href="#">dp3</a>
     <a class="dropdown-item" name="tryme" href="#">dp4</a>
     <a class="dropdown-item" name="tryme" href="#">dp5</a>
    </div>                           
  </div>
</div>

然后这里是一些 JQUERY

$(document).on('click',"a[name='tryme']", function(){
var value = $(this).html();
  $.ajax({
          url: 'query.php',
          type: 'POST',
          data: {value: value},
          success: function(data) {
                    Alert('Success');
                  }
       });
  });
});

查询.php

if(isset($_POST["value"])){
 //code goes here
}

如果您想要https://jsfiddle.net/dtku1rs9/，请在此处尝试