Re: [xsl] as efficient as possible...

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

Corey,

This is a natural for keys. It is bound to speed up processing.

> I am trying to get a list of the products that are listed in the <products>
> node that are also in the <commission/product> node

<xsl:key name="products-by-id" match="products/product" use="id"/>

-- then when your context node is <client>,

select="key('products-by-id', commission/product/productId)"

should work. (Change the XPath to adjust if you need it from a different 
context node.)

(Why is the product ID called 'id' in one place but 'productId' in another? 
Should work anyhow.)

I hope that helps. Any good guide to XSLT should explain keys.

Cheers,
Wendell

At 01:06 PM 7/2/2003, you wrote:
> XML
> ---
> <earners>
>    <earner>
>       <id />
>       <name />
>    </earner>
> </earners>
> <clients>
>    <client>
>    <id>
>    <name>
>    <commissions>
>       <commission>
>          <id />
>          <earnerId />
>          <product>
>            <productId />
>            <rate />
>          </prodcut>
>       </commission>
>     </commissions>
>     <products>
>        <product>
>           <id />
>           <name />
>        </product>
>     </products>
>     </client>
> </clients>
>
>
> XSL
> ---
>
> ./products/product[id=../../commissions/commission[earnerId=$earnerId]/produ
> ct/productId]
>
>
> Problem
> -------
> I am trying to get a list of the products that are listed in the <products>
> node that are also in the <commission/product> node... The way I have above
> works.. but it seems to be very slow... I am working with a large amount of
> data... but I was curious if there was a better way?   $earnerId would be
> the current earner I am trying to calculate totals for... Thanks very much
> for any help!!


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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
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