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Re: [xsl] Getting node w/ lowest attribute value
Subject: Re: [xsl] Getting node w/ lowest attribute value From: "Mukul Gandhi" <gandhi.mukul@xxxxxxxxx> Date: Fri, 15 Aug 2008 23:39:50 +0530 |
Please try this, addresses/address[(@IsActive = 'true') and (@NumOrder = min(../address[@IsActive = 'true']/@NumOrder))][1] This uses the XPath 2.0 function, 'min'. On Fri, Aug 15, 2008 at 11:22 PM, Bordeman, Chris <Chris.Bordeman@xxxxxxxxxxxxxxxxx> wrote: > Hi all. > > I have some nodes like: > > <addresses> > <address IsActive="false" NumOrder=1>[...]</address> > <address IsActive="true" NumOrder=3>[...]</address> > <address IsActive="true" NumOrder=2>[...]</address> > </addresses> > > How do I get the first address node where IsActive=true AND has the > lowest value for the NumOrder attribute? > > In the above case I'd want the 3rd address node (IsActive="true" and > NumOrder=2). > > Any assistance would be appreciated. Thanks. -- Regards, Mukul Gandhi
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