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Answer 1

+1 A:

You can do it if you have a single containing element.

$("<div>").append(data).hide().appendTo("#posts").slideDown();

Let me give you an example:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"&gt;
<html>
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.0/jquery.min.js"&gt;&lt;/script&gt;
<script type="text/javascript">
var content = '<p>four</p><p>five</p><p>six</p>';
$(function() {
  $("#add").click(function() {
    $("<div>").append(content).hide().appendTo("#container").slideDown(1000, function() {
      $(this).contents().unwrap();
    });
  });
});
</script>
<style type="text/css">
div, p { padding: 0; margin: 0; border: 0 none; }
#container { background: yellow; border: 1px solid black; }
</style>
</head>
<body>
<div id="container">
<p>one</p>
<p>two</p>
<p>three</p>
</div>
<input id="add" type="button" value="Add Content">
</body>
</html>

Here the content is not one block, it's three (paragraphs). If it was one block you wouldn't need to wrap it like this.

cletus 2010-03-14 00:58:00

but it is not singe. There ares some posts in it already, and several new may be added sequentially.

Qiao 2010-03-14 01:03:16

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jQuery append while slideToggle

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