Subject: RE: [xsl] Using position() with sorted node From: Jenny Simpson <simpson@xxxxxxxxxxx> Date: Wed, 14 Mar 2001 10:35:55 -0700 |
Thanks for the suggestions on this problem. Instead of using position() to identify sorted nodes, I ended up just assigning unique numerical identifiers (in this case the date stamp) to each node and accessing them that way. This makes the processing/sorted order unimportant. Unsurprisingly, this is a simple solution for what seemed like a complicated problem. Thanks. Jenny Simpson SCI Institute > On Wed, 14 Mar 2001, Michael Kay wrote: > > > > position() does return the position of the current node in > > > the current node > > > list (which is created using an <xsl:apply-templates> or > > > <xsl:for-each> with > > > or without any <xsl:sort> elements.) Unfortunately, sorting > > > the node list > > > makes accessing the preceding nodes rather tricky since the > > > preceding and > > > preceding-sibling axes work on the document and not the > > > sorted node list. > > > > I think the only way of accessing the preceding node in sorted order is to > > create a sorted copy of the original data and then process this using the > > node-set() extension function. Either that, or find a different solution to > > the requirement. > > > > Mike Kay > > Software AG > > > > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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