Question

我有一个通过此链接激活的浮动图层：

<a href="javascript:ToggleFloatingLayer('FloatingLayer',1);"> BUTTON </a>

这是浮动层：

<div id="FloatingLayer">
           <div id="closeX"> <a href="#" onClick="ToggleFloatingLayer('FloatingLayer',0);return false">x</a>
           </div>

剧本：

<script language="JavaScript1.2">

      function ToggleFloatingLayer(DivID, iState) // 1 visible, 0 hidden
      {
        if(document.layers)    //NN4+
        {
           document.layers[DivID].visibility = iState ? "show" : "hide";
        }
        else if(document.getElementById)      //gecko(NN6) + IE 5+
        {
            var obj = document.getElementById(DivID);
            obj.style.visibility = iState ? "visible" : "hidden";
        }
        else if(document.all)   // IE 4
        {
            document.all[DivID].style.visibility = iState ? "visible" : "hidden";
        }
      }
    </script>

我想要＆＃34; BUTTON＆＃34;打开并关闭此浮动图层。所以它在同一个链接中打开和关闭。但是现在我只能通过＆＃34; closeX＆＃34; X。我该怎么办？

Answer 1

jQuery是标准的跨浏览器和功能丰富的JavaScript库在您的应用程序中学习和使用jQuery是所有业务应用程序中最好的以下是api和倾斜网站的链接

代码段在您的下方

<!DOCTYPE html>
<html>

<head>
  <script src="//cdnjs.cloudflare.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
  <script>
    function toggleFloatingLayer(divID) {
      $("#" + divID).toggle();//its only single line to manage toggling for all browsers
    }
  </script>
</head>

<body>

  <a href="#" onclick="toggleFloatingLayer('FloatingLayer')"> BUTTON </a>

  <div id="FloatingLayer" style="display:none;border:solid 2px silver;">
    <div id="closeX" style="background:#efefef"> <a href="#" onClick="toggleFloatingLayer('FloatingLayer')">x</a>
    </div>
    <div>
      Test content of Floating layer
    </div>
  </div>

</body>

</html>

打开和关闭浮动层相同的链接？

1 个答案: