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AW: Re: AW: Variables and command line params

2004-11-09 06:17:01
Sure, thank you very much for the correction.
Got a little bit mixed up...

Wbr,
Roman

-----Ursprüngliche Nachricht-----
Von: xptm(_at_)sapo(_dot_)pt [mailto:xptm(_at_)sapo(_dot_)pt]
Gesendet: Dienstag, 9. November 2004 13:26
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Betreff: [Maybe Spam] Re: AW: [xsl] Variables and command line params

You mean,

<xsl:param name="filename"/>

Citando Huditsch Roman <Roman(_dot_)Huditsch(_at_)lexisnexis(_dot_)at>:

Hi Kevin,

Have you declared your second variable "filename", too?

<xslvariable name="filename"/>

Wbr,
Roman

-----Ursprüngliche Nachricht-----
Von: Kevin Bird 
[mailto:kevin(_dot_)bird(_at_)matrixdigitaldata(_dot_)co(_dot_)uk]
Gesendet: Dienstag, 9. November 2004 12:55
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Betreff: [xsl] Variables and command line params

Hi

I want to pass a second filename as a parameter from the command line.


The below variable declaration works in my stylesheet when the
filename
is hardcoded.

<xsl:variable name="list2" select="doc('filename.xml')//supp"/>

But when I change to...

<xsl:variable name="list2" select="doc($filename)//supp"/>

...and add

          filename="filename.xml"

to the command line option, I get an error - $filename has not been
declared.

Can someone please point out what I am doing wrong.

Thanks.

--
Kevin

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