ah yes, this makes sense now.

thanks for the help,
mike.

"Marrow" <marrow@s...> wrote in message news:<shf8d.126$9H6.28@n...>...
> Hi,
> 
> Try something like...
> 
> <?xml version="1.0"?>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
> <xsl:output method="xml" indent="yes"/>
> 
> <xsl:key name="kLookupB" match="b[@j='aaaa']" use="@i"/>
> 
> <xsl:template match="a">
>   <x>
>     <xsl:for-each select="c[key('kLookupB',@i)]">
>       <y i="{position()}" j="{@j}"/>
>     </xsl:for-each>
>   </x>
> </xsl:template>
> </xsl:stylesheet>
> 
> 
> HTH
> Marrow
> http://www.marrowsoft.com - home of Xselerator (XSLT IDE and debugger)
> http://www.topxml.com/Xselerator
> 
> 
> "Mike Dickens" <mike@p...> wrote in message
> news:721b66d8.0410040823.33819f72@p......
> > hi,
> > i have a query regarding numbering one set of elements as filtered by
>  another set.
> > eg if i have
> >
> > <?xml version="1.0" encoding="ISO-8859-1"?>
> > <a>
> >   <b i="1" j="aaaa"/>
> >   <b i="2" j="bbbb"/>
> >   <b i="3" j="aaaa"/>
> >   <c i="1" j="xxxx"/>
> >   <c i="2" j="yyyy"/>
> >   <c i="3" j="zzzz"/>
> > </a>
> >
> > and i want a numbered c/@j with the same @i as a, where a/@j = "aaaa", ie
>  i get
> >
> > <x>
> >  <y i="1" j="xxxx"/>
> >  <y i="2" j="zzzz"/>
> > </x>
> >
> > (note the @i sequence is 1,2 not 1,3)
> >
> > what would the xslt look like to do this?
> >
> > thanks,
> > mike.