Subject: Re: [xsl] Filter by id and date From: Wolfgang Laun <wolfgang.laun@xxxxxxxxx> Date: Mon, 5 Sep 2011 12:00:46 +0200 |
fn:max() applied to //item[@id=$idref]/@date gives you the maximum date of the sequence with that specific id, and you know how to use that in the <xslvalue-of ...>. This implies that date values are unique. -W On 5 September 2011 11:36, Emma Burrows <Emma.Burrows@xxxxxxxxxxx> wrote: > 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!
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