Need som XSL assistance

[stephen1313]

I would like to create an xsl file that takes XML info and displays it in HTML.XML format is:<Entry> <Author> <FirstName>joe</FirstName> <LastName>smith</LastName> </Author> <Author> <FirstName>bill</FirstName> <LastName>smith</LastName> </Author> <Author> <FirstName>dave</FirstName> <LastName>smith</LastName> </Author> <Author> <FirstName>mary</FirstName> <LastName>smith</LastName> </Author> <Title>cool writings</Title> <Booktitle>XML guideline</Booktitle> <Year>3000</Year> <Type>Journal</Type> <Topic>SEC</Topic> <Download>book1.doc</Download> </Entry>I want the output to be sorted by year (in an ordered table (decending by year and month).I would also like to have a select statement that works off of the year so I can group by year (I.E. 2010, 2009, 2008 etc.)Any help is appreciated.

[Martin Honnen]

Grouping can be done in XSLT 2.0 with for-each-group e.g.

<xsl:for-each-group select="//Entry" group-by="Year">

To sort do

<xsl:for-each-group select="//Entry" group-by="Year">  <xsl:sort select="Year" data-type="number" order="descending"/>

[bitdowner]

Hello MartinYou can use RenderX XEP for generating HTML from XML+XSLRegards

[Martin Honnen]

The documentation of RenderX XEP says it is an XSL-FO processor implemented in Java so you might be using some additional XSLT processor implemented in Java. That way you could switch the XSLT processor to Saxon 9 and use XSLT 2.0 code I posted earlier. I don't know how to use Saxon with XEP, maybe they have a forum where you can ask whether/how that is possible.

[bitdowner]

The documentation of RenderX XEP says it is an XSL-FO processor implemented in Java so you might be using some additional XSLT processor implemented in Java. That way you could switch the XSLT processor to Saxon 9 and use XSLT 2.0 code I posted earlier. I don't know how to use Saxon with XEP, maybe they have a forum where you can ask whether/how that is possible.

Hello, It is RenderX forum ... try it forum