RE: Selecting all descendants with no child nodes

Subject: RE: Selecting all descendants with no child nodes
From: "Evan Lenz" <elenz@xxxxxxxxxxx>
Date: Tue, 3 Oct 2000 14:49:28 -0700
try:

//*[not(*)]


* in the context of a predicate (where the expression automatically gets
converted to a boolean) is converted to true if the node-set it returns is
not empty.  Surround it with not() and the expression will return all
leaves.

Evan Lenz
elenz@xxxxxxxxxxx
http://www.xyzfind.com
XYZFind, the search engine *designed* for XML
Download our free beta software: http://www.xyzfind.com/beta


-----Original Message-----
From: owner-xsl-list@xxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxx]On Behalf Of Taras Tielkes
Sent: Tuesday, October 03, 2000 1:25 PM
To: xsl-list@xxxxxxxxxxxxxxxx
Subject: Selecting all descendants with no child nodes


Hi,
(I hope people don't mind a beginner xpath question now and then)

I'm using the XPath expression "//*[count(*)=0]" to locate all "endpoint"
nodes.

Is there any other way to achieve this, an alternative syntax?

Thanks in advance,
Taras



 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


Current Thread