|
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 |
|---|
|
| <- Previous | Index | Next -> |
|---|---|---|
| Re: [xsl] Length of a literal strin, David Carlisle | Thread | Re: [xsl] Finding the current node , Abel Braaksma |
| Re: [xsl] Length of a literal strin, Abel Braaksma | Date | RE: [xsl] Length of a literal strin, Darren Wheatley |
| Month |