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![]() | ![]() | ![]() | Altova Mailing List Archives>Archive Index >microsoft.public.xsl Archive Home >Recent entries >Thread Prev - sorted distinct nodes with xsl [Thread Next] Re: sorted distinct nodes with xslTo: NULL Date: 9/6/2007 5:54:00 AM Search and read about "grouping methods in XSLT" or "Muenchian grouping" In XSLT 2.0 things are easier as there is the <xsl:for-each-group .../> instruction. Cheers, Dimitre Novatchev "Praveen" <praveen@n...> wrote in message news:%23D%23PAQE8HHA.3940@T...... > XML is like this. > <R> > <I U="6" S="Six"/> > <I U="1" S="One"/> > <I U="2" S="Two"/> > <I U="1" S="One"/> > <I U="2" S="Two"/> > <I U="1" S="One"/> > </R> > > What to get all 'I' nodes with unique 'U' and sorted by 'U' > > otherwise want to print only. > > One > Two > Six > > in xsl. > > thanks, > Praveen > | ![]() | ![]() | ![]() |
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