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Re: using parameters in a select
- To: <xsl-list at lists dot mulberrytech dot com>
- Subject: Re: [xsl] using parameters in a select
- From: lwong at neocore dot com
- Date: Tue, 24 Apr 2001 13:48:49 -0600
- Organization: NeoCore
- References: <4621819DAF2BD411A20200508B641F4302B2E813@MAILSERVER>
- Reply-To: xsl-list at lists dot mulberrytech dot com
Actaually I'm a she, not a he :-)
My $servicePlanNames variable does resolve to the entire xpath statement:
Name='Cingular Nation 100' or Name='Cingular Nation 1500'
Hence, I only want to select service plans with these Name values.
<xsl:apply-templates select="ServicePlan[$servicePlanNames]" />
I *thought* the above line would evaluate to:
<xsl:apply-templates select="ServicePlan[Name='Cingular Nation 100' or
Name='Cingular Nation 1500']" />
However, when I run the trace utility, it seems that the transformer
interprets this as:
Line #39, Column #61: apply-templates, test='ServicePlan[$servicePlans]':
It looks like the transformer does not substitute the value of $servicePlans
before doing the test.
Thanks for your helpful suggestions,
Leah
----- Original Message -----
From: "James Eberhardt" <JEberhardt@extend.com>
To: <xsl-list@lists.mulberrytech.com>
Sent: Tuesday, April 24, 2001 1:19 PM
Subject: RE: [xsl] using parameters in a select
>>>> Can I use parameters in the apply-templates select statement?
> No. Check out the FAQ at http://www.dpawson.co.uk/xsl/nono.html#d44e8531
Actually, it's the reverse of what he is asking. The variable that he would
need goes in the <xsl:apply-templates> part, not the <xsl:template> part,
like the FAQ is indicating.
The problem with his code is the XPath statement that he is using resolves
to nothing.
<xsl:apply-templates select="ServicePlan[$servicePlanNames]" />
What is supposed to equal $servicePlanNames ??
If you had a bit of XML like:
<ServicePlan type="default value" />
Then your xsl would have to look like this:
<xsl:apply-templates select="ServicePlan[type=$servicePlanNames]" />
Hope this helps.
James
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