[xsl] RE: Filter by id and date

[Emma Burrows <Emma.Burrows@xxxxxxxxxxx>]

As usual, the solution came to me after I hit send.

normalize-space(//item[@id=$idref][@date=max(//item[@id=$idref]/@date)])

seemed to be enough to do the trick (putting the "//item[@id=$idref]" into a
variable would be an improvement, but that's the general principle). Thanks
for listening to my monologue anyway. :)


-----Original Message-----
From: Emma Burrows [mailto:Emma.Burrows@xxxxxxxxxxx]
Sent: 05 September 2011 10:36
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] Filter by id and date

Hopefully a quick question. Using XSLT 2.0, I am filtering entries in an input
file based on their ids. The only problem is that I've been given an update to
the original input file which includes many duplicates, some of which are
identical and some of which have been updated. For any given id, I need to
look up the item with the most recent date.

So say I have the following input file:

<items>
  <item id='1' date='2011'>Item1</item>
  <item id='2' date='2009'>Item2</item>
  <item id='3' date='2002'>Item3</item>
  <item id='2' date='2011'>Item2</item>
  <item id='3' date='2002'>Item3</item>
</items>

For example, the Xpath in question looks like this right now (long-winded but
more efficient path-finding syntax simplified to // for this example, and
$idref obviously contains a valid id):

<xsl:value-of select="normalize-space(//item[@id=$idref])"/>

What is the best way to filter that to return only the item whose @date is the
highest for all items with @id=$idref?

Thanks!


______________________________________________________________________
This email has been scanned by the MessageLabs Email Security System.
For more information please visit http://www.messagelabs.com/email
______________________________________________________________________