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XSLT with 2 XML input-variables
Added by Anonymous over 16 years ago
Legacy ID: #5163465
Legacy Poster: vertig (vertig)
Hi all, Please find below a concrete issue regarding an XSL Transformation. Please
do NOT hesitate to sent your ideas / feedbacks / comments and ...hopefully solutions.
Many thanks, Regards, Nicolas Guerin I have actually two XML input-variables: the first
variable is a list of friends (ID / Name) and the second variable is a list of IDs. I
would like to use a XslT to get as output-variable a list of friends with only the
friends whose ID are included in the list of IDs (2nd variable) Does anybody knows how
to write the corresponding xslt ? First input-variable: <friendList>
<name> <ID>1</ID>
<Name>Nico</Name> </friend>
<friend> <ID>2</ID>
<Name>Bob</Name> </friend>
<friend> <ID>3</ID>
<Name>John</Name> </friend>
<friend> <ID>4</ID>
<Name>Mike</Name> </friend>
</friendList> 2nd input-variable: <IDList>
<ID>1</ID> <ID>3</ID>
</IDList> Requested output-variable (after transformation):
<friendList> <name> <ID>1</ID>
<Name>Nico</Name> </friend>
<friend> <ID>3</ID>
<Name>John</Name> </friend>
</friendList>
Legacy ID: #5163531
Legacy Poster: Willem van Heerde (wheerde)
Hi Nicolas, Try something like this $var1 is the friendlist and $var2 the id-list):
<friendList> <xsl:for-each
select="$var2/idlist/id"> <xsl:copy-of
select="$var1/friendlist/friend[id = current()] "/>
</xsl:for-each> </friendList> Or even shorter:
<friendList> <xsl:copy-of
select="$var1/friendlist/friend[id = $var2/idlist/id] "/>
</friendList> I did not test it but I think this should work. Willem.
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