The first time you asked this you said "no extensions". The fact that
you got no answers probably means that this is very hard to do without
extensions.
The problem is actually underspecified. What do you mean for two
elements to be "the same". Just the same name? Or the same value as
well?
Here's an XPath 2.0 solution:
select="//yyy[some $x in //xxx satisfies deep-equal(., $x)]"
I'm afraid I'm too busy writing XSLT 2.0 specs right now to spend time
finding a 1.0 solution for you!
Michael Kay
Software AG
home: Michael(_dot_)H(_dot_)Kay(_at_)ntlworld(_dot_)com
work: Michael(_dot_)Kay(_at_)softwareag(_dot_)com
-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of
Leena Kulkarni
Sent: 10 April 2003 14:45
To: XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] nodes having same values in same order
Hello,
If I have a structure like the following -
<xxx>
<pp></pp>
<qq></qq>
<rr></rr>
</xxx>
<xxx>
<pp></pp>
<qq></qq>
</xxx>
<yyy>
<pp></pp>
<qq></qq>
</yyy>
<yyy>
<pp></pp>
<qq></qq>
<rr></rr>
</yyy>
How do I find out the yyy elements that have exactly
the same elements(same order and count) as in xxx?
Thanks
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