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The actual template does more than copying; it uses recursion to group sections started by headlines at different levels defined by its attributes. I got it from this post:
http://www.biglist.com/lists/lists.mulberrytech.com/xsl-list/archives/200910/msg00228.html
The function works really well (except for cases when a headline level is skipped). So if its possible to keep copying the nodes it would be great; otherwise I'll have to fix the function (I'm thinking of expanding the recursion into four xls:for-each-group statements - it's not allowed to go deeper anyway).
I'll do sone further testing once I get back to the computer.
Den 23. okt. 2009 kl. 15.20 skrev "Michael Kay" <mike@xxxxxxxxxxxx>:
Re: [xsl] XPath problem with getting all ancestors
Subject: Re: [xsl] XPath problem with getting all ancestors From: Jostein Austvik Jacobsen <josteinaj@xxxxxxxxx> Date: Fri, 23 Oct 2009 16:25:25 +0200 |
The actual template does more than copying; it uses recursion to group sections started by headlines at different levels defined by its attributes. I got it from this post:
http://www.biglist.com/lists/lists.mulberrytech.com/xsl-list/archives/200910/msg00228.html
The function works really well (except for cases when a headline level is skipped). So if its possible to keep copying the nodes it would be great; otherwise I'll have to fix the function (I'm thinking of expanding the recursion into four xls:for-each-group statements - it's not allowed to go deeper anyway).
I'll do sone further testing once I get back to the computer.
Regards Jostein
Den 23. okt. 2009 kl. 15.20 skrev "Michael Kay" <mike@xxxxxxxxxxxx>:
<xsl:template match="article"> <xsl:variable name="v1"> <xsl:copy-of select="content/node()"/> </xsl:variable> <xsl:sequence select="mf:group($v1/node())"/> </xsl:template>
As I suspected, you are copying the nodes, and the copies do not have the
same ancestors.
It's not clear to me WHY you are copying them - it seems quite unnecessary.
Regards,
Michael Kay http://www.saxonica.com/ http://twitter.com/michaelhkay
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