Re: [jQuery] simple code not working

2009-12-21 Thread Charlie Griefer
couple of debugging steps I'd take: 1) remove the .getJSON() call and just alert('foo'), to ensure that the change event is being triggered as you expect 2) install Firebug in Firefox and check whether or not there are errors in the php document. On Mon, Dec 21, 2009 at 8:16 AM, kikloo wrote: >

[jQuery] simple code not working

2009-12-21 Thread kikloo
hi i have this simple code: $(function() { $("select#ctlByName").change(function() { cid = $(this).val(); $.getJSON("ajax.php",{cid:cid}, function(json) { alert("Data Loaded: " + json); }) }) }) and its not

[jQuery] Simple Code, not working.

2008-11-09 Thread Jason
var lawls = 1; $('#next').click(function () { $('.item').hide(); $('.item:eq(lawls)').fadeIn(); ++lawls; }); I have multiple divs, that I would like to sequentially appear when a button is clicked. My current code isn't working. This will work: $('#next').click(function