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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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