Subject: [xsl] RE: Using a reference in a sort From: "Jesse M. Heines" <heines@xxxxxxxxxx> Date: Tue, 29 Jul 2003 19:37:51 -0400 |
> Try <xsl:sort > select="document($filePeople)/people/person[@id=current()/person/@id]/@l ast" > order="descending"/> instead Thanks for your reply, but I'm sorry to report that that doesn't work either. The context of current() is still the node in the other XML file. The documentation I have says that current() is the same as ".", which I tried and does not help, either. More suggestions, please! Jesse XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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