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Hi,
Thanks again to everyone who has replied to these messages. I had no idea how hard this would be. It seemed so simple when I explained it to the client ;-)
I'm looking for an XPath expression that will find all elements whose ancestors are the same. By "same" I mean their @name is the same at each level of ancestry (and they appear in the same order).
For example, this expression would return a count of 2 for the following XML when filtered on the "daughter" folder (or "mother" or "grandfather" for that matter).
I know it's going to be something like this, but I just can't get my head around it!
<xsl:copy-of select="/root//folder[@name = following-sibling::*/@name and ancestor-or-self::*/@name = following-sibling::*/@name/ancestor-or-self::*/@name" />
Thanks in advance!
Ted
[xsl] Find elements whose ancestors are the same
Subject: [xsl] Find elements whose ancestors are the same From: Gustave Stresen-Reuter <tedmasterweb@xxxxxxx> Date: Mon, 26 Sep 2005 21:38:20 +0100 |
Hi,
Thanks again to everyone who has replied to these messages. I had no idea how hard this would be. It seemed so simple when I explained it to the client ;-)
I'm looking for an XPath expression that will find all elements whose ancestors are the same. By "same" I mean their @name is the same at each level of ancestry (and they appear in the same order).
For example, this expression would return a count of 2 for the following XML when filtered on the "daughter" folder (or "mother" or "grandfather" for that matter).
<root> <folder name="grandfather"> <folder name="mother"> <folder name="daughter" /> </folder> </folder> <folder name="grandfather"> <folder name="mother"> <folder name="daughter" /> </folder> </folder> <folder name="grandmother"> <folder name="father"> <folder name="son" /> </folder> </folder> </root>
I know it's going to be something like this, but I just can't get my head around it!
<xsl:copy-of select="/root//folder[@name = following-sibling::*/@name and ancestor-or-self::*/@name = following-sibling::*/@name/ancestor-or-self::*/@name" />
Thanks in advance!
Ted
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