jQuery slideToggle函数的问题

Question

Please help me find the problem with this code. I am trying to create a simple slideToggle element! I tried this first:

$("..").slideToggle(..,..,..)

My current code looks like this:

My code:

 $(document).ready(function () { $("#Layer13copy2").click(function () { $("#toggle").animate({ height: 'toggle' }); }); }); 

 #Layer13copy2 { left: 13.85%; top: 398px; position: absolute; width: 892px; height: 37px; z-index:5; background:url('../images/slidet.png') 45%; background-repeat:no-repeat; background-size:101.4% 101%; } #toggle { z-index: 4; background-color: #3a97b1; display: block; left: 13.85%; top: 39.37%; position: absolute; width: 71.95%; height: 17%; } 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script> <div id="Layer13copy2"></div> <div id="toggle"></div> 

Answer 1

  $(document).ready(function(){ $('#toggle').animate({ height : 'toggle' }); $("#Layer13copy2").click( function(){$('#toggle').animate({ height : 'toggle' }); });}); 

  #Layer13copy2 { position: absolute; width: 292px; height: 37px; background-color:blue; } #toggle{ background-color: #3a97b1; top: 35px; position: absolute; width: 100%; height: 100px; } 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script> <div id="Layer13copy2" > <div id="toggle" marginwidth="0"></div> </div>