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Assuming 'foo' elements can have a 'bar' and a 'baz' attribute and you want to process them sorted on @bar if present, otherwise on @baz then the following XSLT 2.0
does that nicely I think.
How would you translate that to XSLT 1.0? I can only think of a temporary tree (created with a node-set extension function) where you create a new attribute merging the two attributes first, then process the temporary tree sorted on the new attribute.
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[xsl] xsl:sort with XSLT 2.0 translated to XSLT 1.0
Subject: [xsl] xsl:sort with XSLT 2.0 translated to XSLT 1.0 From: Martin Honnen <Martin.Honnen@xxxxxx> Date: Thu, 19 Jun 2008 15:26:47 +0200 |
Assuming 'foo' elements can have a 'bar' and a 'baz' attribute and you want to process them sorted on @bar if present, otherwise on @baz then the following XSLT 2.0
<xsl:apply-templates select="foo"> <xsl:sort select="(@bar, @baz)[1]"/> </xsl:apply-templates>
does that nicely I think.
How would you translate that to XSLT 1.0? I can only think of a temporary tree (created with a node-set extension function) where you create a new attribute merging the two attributes first, then process the temporary tree sorted on the new attribute.
--
Martin Honnen http://JavaScript.FAQTs.com/
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