[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: Selecting all descendants with no child nodes
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
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