Subject: Re: [xsl] using parameters in a select From: lwong@xxxxxxxxxxx Date: Tue, 24 Apr 2001 13:48:49 -0600 |
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 $servicePlans before doing the test. Thanks for your helpful suggestions, Leah ----- Original Message ----- From: "James Eberhardt" <JEberhardt@xxxxxxxxxx> To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx> Sent: Tuesday, April 24, 2001 1:19 PM Subject: RE: [xsl] using parameters in a select >>>> Can I use parameters in the apply-templates select statement? > No. Check out the FAQ at http://www.dpawson.co.uk/xsl/nono.html#d44e8531 Actually, it's the reverse of what he is asking. The variable that he would need goes in the <xsl:apply-templates> part, not the <xsl:template> part, like the FAQ is indicating. The problem with his code is the XPath statement that he is using resolves to nothing. <xsl:apply-templates select="ServicePlan[$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 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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