At 2011-10-23 17:58 -0700, Mark wrote:
Hi Ken,
I removed the [1] at the end to get all of the returns, and for the
second example a '50' is returned, the @denomination[5] and the
default [0]. It is clear that the @souvenir-sheet value is ignored
in both cases. I am using XSLT 2, so it always collapses. Is there
something incorrect with
../Location[name(@*)=name(current()/@*)]
It seems correct to me.
You are addressing the element and I copied that without thinking. I
think you only want the attribute:
../Location/@*[name(.)=name(current()/@*)]
. . . . . . . . . . Ken
Mark
-----Original Message----- From: Mark
Sent: Sunday, October 23, 2011 5:39 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Can a single XPath statement duplicate the
functionality of this verbose statement?
Hi Ken,
Yes, only one attribute to either element.
You said:
xsl:copy-of select="( ../Location[name(@*)=name(current()/@*)] ,
../Location/@denomination ,
'0' )[1]"/>
but I may have explained my needs incorrectly. Your statement works on the
first example below [uses @denomination] but returns nothing for the second
example; there it should return the value of the <Locations
@souvenir-sheet>, i.e., '1'.
Thanks,
Mark
<Stamp>
<Formats souvenir-sheet="2895"/>
<Location denomination="1"/>
</Stamp>
<Stamp>
<Formats souvenir-sheet="2896"/>
<Location denomination="5"/>
<Location souvenir-sheet="1"/>
</Stamp>
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