如何根据值显示警报弹出窗口

时间:2017-06-22 07:43:15

标签: javascript popup alert

我有一个下拉列表,例如nothing, name1, name2, name3, name4 and etc,...当我选择nothing时,只有警告弹出窗口应显示当我选择其他人时弹出窗口不应显示的位置。

有什么想法吗?

1 个答案:

答案 0 :(得分: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>