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Posted to fop-users@xmlgraphics.apache.org by bonekrusher <dj...@yahoo.com> on 2009/06/16 14:27:02 UTC

Command Line xsl paramters

Hi, Using FOP 20.5 command line, I am trYing to pass a xsl parameter to my
stylesheet but I think my syntax is incorrect. Is this correct?

-param file-path testtest

EXAMPLE:

java -Dfile.path=%XPATH% -Djava.endorsed.dirs=..\ -cp
.;%LOCALCLASSPATH%;%FOPHOME%fop.jar -Xms256m -Xmx1056m
org.apache.fop.apps.Fop  -xml %1 -xsl %xslfile% -pdf %1.pdf  -param
output-graphics %outgraphics% -param file-path testtest
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Re: Command Line xsl paramters

Posted by bonekrusher <dj...@yahoo.com>.
Thanks again Andreas - you're always a great help. As of now, I am stuck with
20.5 for this project. I will perform the stages separately.  

Thx

Phil




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Re: Command Line xsl paramters

Posted by Andreas Delmelle <an...@telenet.be>.
On 16 Jun 2009, at 14:27, bonekrusher wrote:

Hi Phil

> Hi, Using FOP 20.5 command line, I am trYing to pass a xsl parameter  
> to my
> stylesheet but I think my syntax is incorrect. Is this correct?
>
> -param file-path testtest

IIRC, FOP 0.20.5 did not yet offer the possibility of setting XSLT  
parameters via the command-line. If you're stuck on that version, I  
see no other options than performing the XSLT stage separately, and  
feeding the FO output to FOP.

Regards

Andreas

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