如何根据值显示警报弹出窗口

Question

我有一个下拉列表，例如 nothing, name1, name2, name3, name4 and etc,... 当我 select nothing 时，只有警报弹出窗口应该显示，而当我 select 其他人时弹出窗口不应该显示。

有什么想法吗？

Answer 1

$(document).ready(function(){
     $("#dropdown_change").change(function(){
    if(document.getElementById("dropdown_change").value == "nothing"){
    alert("nothing");
    $("#popup").css("display", "block");
    }
     });
   });

#popup{
    position: fixed;
    margin:auto;
    background: rgba(0,0,0,0.5);
    display: none;
    top: 0;
    left: 0;
    bottom:0;
    right:0;
    width: 300px;
    height: 200px;
    border: 1px solid #000;
    border-radius: 5px;
    padding: 5px;
    color: #fff;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<!DOCTYPE html>
<body>
<form id="myform">

   Select a value from the list:
 
   <select id="dropdown_change">
      <option value="name1">name1</option> 
      <option value="nothing" id="open-popup">nothing</option>

      <option value="name2">name2</option>
 
      <option value="name3">name3</option>
  
      <option value="name4">name4</option>
   
 
</select>
</form>
<div id="popup"> POP UP</div>
</body>

如何根据值显示警报弹出窗口

how to display alert popup based on a value

javascript

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