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#1
May 29th, 2003, 12:04 AM
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Node counting in xsl
Given an xml doc like so:
Code:
<root> <a /> <a /> <b /> <a /> <b /> </root> For a given transform of b, I want to be aware of which node it being transformed. So, the xsl template may appear like this: Code:
<xsl:template match="a"> This is an a element. </xsl:template> <xsl:template match="b"> This is b element number: <!-- What? --> </xsl:template> That is, the output would be something like: Code:
This is an a element. This is an a element. This is b element number: 1 This is an a element. This is b element number: 2 That is, the position() function can't be used, since the b nodes are intermixed with the a nodes. Does anyone know a way to do this? -Michael
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#2
May 29th, 2003, 12:36 PM
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Thanks. I thought of that though, and I don't see how I can use an iterative loop like that while still preserving the relative order of the a and b nodes.
That is, the loop can easily iterate through all b elements, returning them in order. But I want to be able to preserve the ordering of the original children (both a and b) of the root node, while counting the b nodes.
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#3
May 29th, 2003, 01:16 PM
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Hah, excellent. Thanks echolalia, I'll give that a whirl tonight, it should easily be adaptable to the project I'm working on.
Cheers! -Michael
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