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Re: Testing position of parent
>>>>> "Kay" == Kay Michael <Michael.Kay@icl.com> writes:
>> if the parent is an <LG> which has first position among its
>> siblings, then do X, otherwise do Y
Kay> Try test="parent::lg[not(preceding-sibling::*)]"
>>
>> Here's what I thought would be the expression to accomplish that:
>>
>> <xsl:when test="parent::lg[position()=1]">
Kay> This looks at its position among all the parent::lg elements of the current
Kay> node
I always find it interesting to compare your responses with what David
Carlisle says. Considering how obtuse test and path expressions can
be to newbies (which I still count myself), it's amazing how often
your two answers are almost identical. However, on this question I
notice that your suggestion is slightly different from David's
suggestion. I believe they will return the same result, but I'm
wondering if there's any tradeoffs to each approach.
You said: test="parent::lg[not(preceding-sibling::*)]"
David C. said: test="not(parent::lg[preceding-sibling::*])"
Will there be any performance or other tradeoffs to these two
expressions? Are there any general guidelines for understanding the
performance differences between two expressions like these?
--
===============================================================================
David M. Karr ; dkarr@tcsi.com ; w:(425)487-8312 ; TCSI & Best Consulting
Software Engineer ; Unix/Java/C++/X ; BrainBench CJ12P (#12004)
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