Re: Select Deepest Node only in Iteration
A problem, reversing (descending) works, but selecting the top of the list does not work. I have been testing by outputting some numbers from the following 2 functions. <a><xsl:value-of select="count(ancestor::node())"/></a> <b><xsl:value-of select="position()"/></b> This the output I am getting for the first item in the list: <a>5</a> <b>74</b> So the position() function does not work in my case. That is why I think I need to match in an if statement with count(ancestor::node()), e.g. <xsl:if test="count(ancestor::node())=5"> Then I am getting the top of the list, a problem is however that I am reading multiple documents and match with multiple keywords. So I cannot set it adhoc this way. What I did find out is that the first item in the reversed list is the highest number for each leaf. The only question is: is there a function to replace the "5" in the if test? I hope you understand my problem. Thanks! David Carlisle wrote: > xsl:sort has to be the first child of the xsl:for-each, so move it up a > couple of lines. > > David > > ________________________________________________________________________ > The Numerical Algorithms Group Ltd is a company registered in England > and Wales with company number 1249803. The registered office is: > Wilkinson House, Jordan Hill Road, Oxford OX2 8DR, United Kingdom. > > This e-mail has been scanned for all viruses by Star. The service is > powered by MessageLabs. > ________________________________________________________________________
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