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RE: XALAN - obtaining the -XSL stylesheet name?
- To: <xsl-list at lists dot mulberrytech dot com>
- Subject: RE: [xsl] XALAN - obtaining the -XSL stylesheet name?
- From: "Tim Watts" <timw at 3d3 dot com>
- Date: Thu, 29 Mar 2001 15:30:59 +1000
- Reply-To: xsl-list at lists dot mulberrytech dot com
I don't know of any way of deriving the param from the file name using XSLT.
Creating a stylesheet param with the url argument of pagePopUp will be the
simplest solution to your problem, unless of course someone else on the
xsl-list knows if what you are asking is possible, or you are willing to use
another technology to extract the url.
Tim Watts
-----Original Message-----
From: owner-xsl-list@lists.mulberrytech.com
[mailto:owner-xsl-list@lists.mulberrytech.com]On Behalf Of Robert
Nicholson
Sent: Thursday, 29 March 2001 3:06 PM
To: xsl-list@lists.mulberrytech.com
Subject: RE: [xsl] XALAN - obtaining the -XSL stylesheet name?
You have created a separate stylesheet parameter for this.
I can do this and will probably have to but since what I want to generate is
derived from the stylesheet name I didn't want to have to create the
parameter if there was someway of knowing the stylesheet that was passed.
I think I understand your example here. You are passing a stylesheet
parameter for your title value. That's a little different from what I was
asking I think.
> -----Original Message-----
> From: owner-xsl-list@lists.mulberrytech.com
> [mailto:owner-xsl-list@lists.mulberrytech.com]On Behalf Of Tim Watts
> Sent: Wednesday, March 28, 2001 8:43 PM
> To: xsl-list@lists.mulberrytech.com
> Subject: RE: [xsl] XALAN - obtaining the -XSL stylesheet name?
>
>
> We use something like so...
>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
> <xsl:include href="../document.xsl" />
> <xsl:include href="../common.xsl" />
> <xsl:param name="title">Edit Settings Wizard</xsl:param>
> <xsl:template name="body">
> ...
> </xsl:template>
> </xsl:stylesheet>
>
> with the document.xsl containing
>
> <?xml version="1.0" encoding="utf-8"?>
> <xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
> <xsl:template match="document">
> <html lang="{language/html-language}"
dir="{language/html-text-direction}">
> <head>
> <title><xsl:value-of select="$title"/></title>
> </head>
> <body>
> ...
> <xsl:call-template name="body"/>
> ...
> </body>
> </html>
> </xsl:template>
> </xsl:stylesheet>
>
> This passes the param "title" to the html for use in the external
> stylesheet, simular to how you want to pass a param to your external
> stylesheet.
>
> This can be directly duplicated into your style sheet in the same way. The
> param is just declared in the first stylesheet and can be used in the
> external one.
>
> best wishes,
>
> Tim Watts
>
>
> -----Original Message-----
> From: owner-xsl-list@lists.mulberrytech.com
> [mailto:owner-xsl-list@lists.mulberrytech.com]On Behalf Of Robert
> Nicholson
> Sent: Thursday, 29 March 2001 2:20 PM
> To: xsl-list@lists.mulberrytech.com
> Subject: RE: [xsl] XALAN - obtaining the -XSL stylesheet name?
>
> I want the url argument of pagePopUp of the resulting html to be derived
> from the stylesheet that I'm passing as -XSL to XALAN which is the same
> stylesheet that's importing the common stylesheet.
>
>
> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
>
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