Re: xsl:if giving me "Invalid type" error

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Subject: Re: xsl:if giving me "Invalid type" error
From: Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>
Date: Tue, 15 Apr 2003 16:11:58 -0400

At 02:56 PM 4/15/2003, you wrote:

> How have you declared $firstDiv?

It is a string. :(

...

I was trying to store the name of whatever the first
top-level <div*> was. Maybe

<xsl:variable name="firstDiv">
  <xsl copy-of
select="(div[1]|div0[1]|div1[1]|divGen[1])[1]"/>
</xsl:variable>

would work better for what I need to do (which is to
keep track of that first top-level <div*>)?

Not in XSLT 1.0 it won't. Unless you're willing to use a node-set() extension function on it, it's a Result Tree Fragment and can't be traversed.

This is why I suggested doing it from a stronger key declaration instead. If you need to do this kind of thing, you have to get devious. For example, use generate-id() with a key like

<xsl:key name="elems-by-genid" match="*" use="generate-id()"/>

to retrieve nodes by unique identifiers. (That is, bind the generated id to the variable and then use the key to retrieve the node.)

But this shouldn't be necessary.

Cheers,
Wendell

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Current Thread
Re: xsl:if giving me "Invalid type" error, (continued) James J. Ramsey - Tue, 15 Apr 2003 14:51:35 -0400 (EDT) David Carlisle - Tue, 15 Apr 2003 15:11:16 -0400 (EDT) James J. Ramsey - Tue, 15 Apr 2003 15:11:19 -0400 (EDT) Wendell Piez - Tue, 15 Apr 2003 16:12:59 -0400 (EDT) Wendell Piez - Tue, 15 Apr 2003 16:09:18 -0400 (EDT) <= Passin, Tom - Tue, 15 Apr 2003 15:35:43 -0400 (EDT)

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Re: xsl:if giving me "Invalid, Wendell Piez	Thread	RE: xsl:if giving me "Invalid, Passin, Tom
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