.append in jQuery not working correctly

Question

I am appending div in existing div and then one more div into append div but it not work for me. what am I doing wrong. please help

<head> <script type="text/javascript"> $(function(){ $('a').click(function(){ var cl=$('#free').clone(); var jj=$('.append') var mm=jj.append('<div class="hii"></div>') mm.append(cl) }) }) </script> </head> <body> <div style="background:#F00; width:500px; height:50px" id="free"></div> <a href="#">hide</a> <div class="append"></div> </body>

; is optional in javascript. And above code is working at my end. — Jitendra Pancholi
– Jitendra Pancholi, Commented Aug 22, 2012 at 12:36
You're cloning an element with the id 'free', so you're creating an element that doesn't have an unique identifier. This isn't valid HTML, so avoid doing that. Instead of id="free", do class="free". Also, what is going wrong with this? Can you create a fiddle or something like that which demonstrates this problem? — jakee
– jakee, Commented Aug 22, 2012 at 12:36
Yes its a good practice to always use ;. Its a good programming habit. :) — Jitendra Pancholi
– Jitendra Pancholi, Commented Aug 22, 2012 at 12:50

Danil Speransky · Accepted Answer · 2012-08-22 12:36:08Z

2

jj.append('<div class="hii"></div>') return jj, not new div. So use this:

$(function(){ $('a').click(function(){ var cl = $('#free').clone(); var jj = $('.append'); var mm = $('<div class="hii"></div>'); jj.append(mm); mm.append(cl); }); });

answered Aug 22, 2012 at 12:36

Danil Speransky

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2 Comments

Jitendra Pancholi Over a year ago

@SperanskyDanil & Amit: I can see the same output from both functions.

Jitendra Pancholi Over a year ago

@SperanskyDanil: I have seen that but the question is if his code is wrong, then why its giving the correct output? Could you please explain.

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