Jquery Accordion/Xsl

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So, here is what I am doing:

I am transforming a list of links from an xml with an xsl. With that transformed data I am making it into a jquery accordion list.

Lets start with the code:
Xml (Example not actual):

XML

<links>
  <topic id="one">
    <link>
      <title>
        1
      </title>
      <url>
        http://blahblah.com
      <url>
    </link>
      </url></url>
  </topic>
<links>

But lets say that topic id=one
has three titles (1,2,3)
and topic id=two
has three titles
How would you code the xsl so that it could end up being transformed as an accordion list?

Sorry I do'nt know how to copy in code?

When I use the above code the accordion list end up being
One
1
4
Two
1
4

Instead of:
One
1
2
3
Two
4
5
6
I have my xsl and xml files on google docs, feel free to take a look
https://docs.google.com/leaf?id=0B4bIPmgE94uPZjkxNjUyMzQtMzMyNy00Y2U3LWI1MDAtNzcwODY2ZWU2NzE1&hl=en
https://docs.google.com/leaf?id=0B4bIPmgE94uPNWY5MGE3ZTEtOTk0Zi00MDczLTk4OTEtZmRmNWIzNjM4NGM0&hl=en

Any ideas????

Thanks in advance
:confused:~Brittany Ansteatt~:confused:
*REMOVED MAIL*

Any input would be greatly appreciated

Posted 16-Dec-10 20:25pm

Bansteatt84

Updated 16-Dec-10 20:31pm

JF2015

v2

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JF2015 17-Dec-10 2:31am

Edited to fix links and remove email address. Please never post your email address. You will be notified if someone answers your post.

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lorenkins · Answer 1 · 2010-12-17T12:15:00

You may need to take a closer look at your xml source. The sample has a run of title and url elements as siblings within a single topic/link. I think your xslt will work as you expect if you add some additional <link>....</link> elements to wrap the <title></title><url></url> pairs.

(Note too that within the context of the <xsl:for-each select="/links/topic">, there's no need to repeat that portion of the xpath; you can simply select="link" to render the LI elements.)