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| | | Altova Mailing List Archives>Archive Index >xsl-list Archive Home >Recent entries >Thread Prev - [xsl] Sort multiple xml files with identical keys [Thread Next] RE: [xsl] Sort multiple xml files with identical keysTo: Date: 3/1/2006 2:41:00 PM The Muenchian method doesn't work well across multiple documents, because key() only works within one document at a time. Best approach is to combine the documents first, then do the grouping. Michael Kay http://www.saxonica.com/ > -----Original Message----- > From: Tapio.Niva@xxxxxxxxxxxxxxx [mailto:Tapio.Niva@xxxxxxxxxxxxxxx] > Sent: 01 March 2006 13:51 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: [xsl] Sort multiple xml files with identical keys > > Hello, > I'm using IE6 & MSXML to transform,sort and count 2 separate xml docs, > but the results are not as I expected. > As a beginner, I obviously have understood something wrong, I'm using > the so-called "Muenchian method" here (...wtg for the XSLT > 2.0 to MSXML) > I would be very appreciated if some xslt expert could give me a clue > what I'm doing wrong. > > Below are xml, xsl, results I got and results I wanted : > ------------------------- > net_sample1.xml: > <?xml version="1.0" encoding="UTF-8"?> > <?xml-stylesheet type="text/xsl" href="net_sample.xsl"?> > <List> > <row> > <row_id>A1</row_id> > <row_status>New</row_status> > </row> > <row> > <row_id>A2</row_id> > <row_status>New</row_status> > </row> > <row> > <row_id>A2</row_id> > <row_status>Old</row_status> > </row> > </List> > ------------------------- > net_sample2.xml: > <?xml version="1.0" encoding="UTF-8"?> > <?xml-stylesheet type="text/xsl" href="net_sample.xsl"?> > <List> > <row> > <row_id>A1</row_id> > <row_status>New</row_status> > </row> > <row> > <row_id>A1</row_id> > <row_status>New</row_status> > </row> > <row> > <row_id>A2</row_id> > <row_status>Old</row_status> > </row> > </List> > ------------------------ > net_sample.xsl : > <?xml version="1.0"?> > <xsl:stylesheet version="1.0" > xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > > <xsl:key name="rowkey" match="row" > use="row_id" /> > <xsl:key name="rowkey_old" match="row[row_status='Old']" > use="row_id" /> > <xsl:key name="rowkey_new" match="row[row_status='New']" > use="row_id" /> > <xsl:template match="/"> > <html> > <head></head> > <body> > <xsl:variable name="all_docs" select="document('net_sample1.xml' > )/List/row | document('net_sample2.xml' )/List/row "/> > <table border="0" bgcolor="ivory"> > <xsl:for-each select="$all_docs"> > <xsl:sort select="row_id"/> > <xsl:for-each select="current()[count(. | key('rowkey', > row_id)[1]) = > 1]"> > <tr> > <td><xsl:value-of select="row_id"/></td> > <th>Old rows = </th> > <td><xsl:value-of select="count( key('rowkey_old',row_id) > )"/></td> > <th>New rows = </th> > <td><xsl:value-of select="count( key('rowkey_new',row_id) > )"/></td> > </tr> > </xsl:for-each> > </xsl:for-each> > </table> > </body> > </html> > </xsl:template> > </xsl:stylesheet> > ------------ > Result: > A1 Old rows=0 New rows=1 > A1 Old rows=0 New rows=2 > A2 Old rows=1 New rows=1 > A2 Old rows=1 New rows=0 > ------------------------------------------ > The result I expected to see : > A1 Old rows=0 New rows=3 > A2 Old rows=2 New rows=1 > > Any ideas how to reach the wanted result ? > > Regards, Tapio | |
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