Here's an old, but good, tutorial on xml and jQuery:
http://www.bennadel.com/blog/1054-jQuery-Demo-Working-With-XML-Documents.htm
...or if you want to make the code you have work you can "cheat" and just
change the name of one of the two 'id's in your xml... ;)
Wil
On Wed, Nov 18, 2009 at 1:18 AM, Joe Moore <joe.lynn.mo...@gmail.com> wrote:
> Instead of find, use children?
>
> On Nov 17, 2009 2:56 PM, "Joe" <joecel...@gmail.com> wrote:
>
> I have been banging my head against the wall and searching for answers
> for the past week, so any help or direction would be appreciated. I
> am sure that I am missing something obvious. I have some XML that
> looks like this:
>
> <?xml version="1.0" encoding="UTF-8"?>
> <customMessageList>
> <CustomMessage>
> <id>181</id>
> <name>First Message</name>
> <customMessageImage>
> <id>135</id>
> <name>Sample_img.png</name>
> <contentType>image/png</contentType>
> <modifiedOn>2009-08-25
> 11:25:28.34</modifiedOn>
> </customMessageImage>
> </CustomMessage>
> </customMessageList>
>
>
> And an AJAX call that looks like this:
>
> $.ajax({
> type: "GET",
> url: myURL,
> success: function(xml) {
> $(xml).find('CustomMessage').each(function(){
> var id_text = (this).find('id').text();
> var name_text = $(this).find('name').text();
> var contentType_text =
> $(this).find('contentType').text();
> }); //close each(
> }
> }
> });
>
>
> When i run this code, the id and name variables have all the id/name
> nodes in them. I only want the id/name for the child node. the
> contentType variable works find (but there is only one instance.
>
> Any ideas on how to fix this? I am at my wits end.
>
> thx,
>
> Joe C
>
>
