thx, i tried that. whether i do a find or a children, it still returns both id's. you can try it with a simple app that loads that XML. it's really odd.
i still haven't got a solution. On Nov 18, 2:12 pm, Wil Everts <wileve...@gmail.com> wrote: > Here's an old, but good, tutorial on xml and > jQuery:http://www.bennadel.com/blog/1054-jQuery-Demo-Working-With-XML-Docume... > > ...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
