swapping <div>?
this is my sample code
<div id="id-1"><div>sampledata</div><span>sampledata</span></div>
<div id"id-2"><div>sampledata</div><div> smpledata</div></div>
<div id"id-3"><div>sampledata</div><div> smpledata</div></div>
<div id"id-3"><div>sampledata</div><div> smpledata</div></div>
i'm having the id of开发者_高级运维 only one div(for example id-3) . i want to swap that <div> with next <div>, since there is more than one <div> inside a <div> i cant use a next() .
thanks in advance
Well, select div, then find next and insertAfter next
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.5.0/jquery.js"></script>
<style>
.innderdiv4{
backgrournd: green;
}
.innderdiv3{
background: blue;
}
</style>
<script>
$(function (){
var div = $('#id3');
var next = $(div).next();
if(confirm("Are you sure you want to do it"))
{
$(div).insertAfter(next);
}
})
</script>
<div id="container">
<div id="id3">
<div class="innderdiv3">div 3</div>
<div class="innderdiv3">div 3</div>
</div>
<div id="id4">
<div class="innderdiv4">div 4</div>
<div class="innderdiv4">div 4</div>
</div>
</div>
.next(); does select the very next sibling, not the child div elements. However, to ensure that you select the next sibling that is actually a div, you should use .next("div");.
$("#id-3").before($("#id-3").next("div"));
The code above works with the example you gave as long as you change your last div's id to "id-4".
Or, you can copy and paste this code for a working example:
<html>
<head>
<title>Div Swap Test Page</title>
<style type="text/css">
#id-1,#id-2,#id-3,#id-4 {height: 50px; margin:10px 0; border:solid 1px #000}
#id-3 {border:solid 1px #D00}
#id-4 {border:solid 1px #0D0}
</style>
</head>
<body>
<div id="id-1"><div>This is div #1</div><span>I am first.</span></div>
<div id="id-2"><div>This is div #2</div><span>I am second.</span></div>
<div id="id-3"><div>This is div #3</div><span>I am third.</span></div>
<div id="id-4"><div>This is div #4</div><span>I am last.</span></div>
<p><button id="btn"><span>Swap</span></button></p>
</body>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.5.0/jquery.min.js" type="text/javascript"></script>
<script type="text/javascript">
$(function() {
$("#btn").click(function() {
$("#id-3").before($("#id-3").next("div"));
});
});
</script>
</html>
加载中,请稍侯......
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