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[xsl] Finding the current node type

Subject: [xsl] Finding the current node type
From: "Mark Anderson" <mark.anderson@xxxxxxxxxxxxxxxxxxx>
Date: Wed, 1 Aug 2007 14:36:40 +0100

Thanks Guys

In my application, I will only need to process a few hundred nodes and
about 20-30 of the type where I need to use the choose statement.

I went with the <xsl:if test=self::speaker> option as I still use the
namespace (e.g. <xsl:if test=self::abc:speaker>). Whereas localname
returned the node name without the ns prefix

Thanks for the help

Regards

mark

Current Thread
[xsl] Finding the current node type Mark Anderson - 1 Aug 2007 13:38:33 -0000 <= Abel Braaksma - 1 Aug 2007 14:14:06 -0000 J. Zhang - 1 Aug 2007 15:38:06 -0000 Abel Braaksma - 1 Aug 2007 15:42:47 -0000 Michael Kay - 1 Aug 2007 15:55:47 -0000

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Re: [xsl] Length of a literal strin, David Carlisle	Thread	Re: [xsl] Finding the current node , Abel Braaksma
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