Subject: Re: position() and last() problem
From: David Carlisle <davidc@xxxxxxxxx>
Date: Wed, 8 Jan 2003 16:28:03 GMT
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ancestor::con:view[1]
selects a node list of at most 1 node so position()=1=last() always
so the above would be equivalent to
ancestor::con:view[1][position()=last()]
I thought you had written
ancestor::con:view[1][position()!=last()]
which will always selct an empty node set as position() and last() are
both 1 so never not equal.
However U see you have written
"ancestor::con:view[1][position()]!=last()"
which is legal but compares the numeric value of the data content
of the parent con:view element against the number of nodes in the
current node list at the point the xsl:if is evaluated.
[position()] is short for [position()=position()] which is always
true so the above is
"ancestor::con:view[1]!=last()"
which is
"parent::con:view != last()"
This almost certainly isn't what you want to do.
If I understand your problem you want to know if the parent has a
following sibling
test="parent::*/following-sibling::*"
David
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| Current Thread |
- position() and last() problem
- jfi - Wed, 8 Jan 2003 10:44:55 -0500 (EST)
- David Carlisle - Wed, 8 Jan 2003 11:28:16 -0500 (EST) <=
- Michael Kay - Wed, 8 Jan 2003 13:23:17 -0500 (EST)
- <Possible follow-ups>
- Andrew Welch - Wed, 8 Jan 2003 11:43:53 -0500 (EST)
- jfi - Wed, 8 Jan 2003 12:00:36 -0500 (EST)
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