Re: [xsl] How do I properly define a unix saxon script command line

Subject: Re: [xsl] How do I properly define a unix saxon script command line
From: "Houghton,Andrew houghtoa@xxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Tue, 11 Nov 2014 22:08:26 -0000
This should help:
http://www.saxonica.com/html/documentation/using-xsl/commandline.html

Just put Java in your Unix path or specify the full path name to java in the
command script.

Andrew

From: Catherine Wilbur cwilbur@xxxxxxxxxxx
[mailto:xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx]
Sent: Tuesday, November 11, 2014 16:47
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] How do I properly define a unix saxon script command line

Want to write a command line to call Saxon processor from a unix script line.
Am using Saxon-HE from Unix command line to convert my XML file into a CSV
file so it can be uploaded into a DB Table. We are using the DB table as input
into our AP interface in ERP E1. Had to switch to Saxon processor because
xsltproc does not work with xsl 2.0.

In the command line call I need to specify two classpaths  (Classpath for
JAVA, Classpath for SAXON jar files)

How would I call the saxon processor in a unix script and pass it the
following information
JavaClassPath, SaxonJarClassPath, InputFile, StylesheetFile, OutputFile

So far this is the unix command line I came across on the web.  Read thru page
853-854 of XSLT 2.0 3rd Edition Programmer's Reference and it does not quite
give me what I am looking for.  Where would I find all the Saxon command line
options
Want to do saxon conversion of XML file within a unix shell script.  We have
Java on Unix box.  Loaded Saxon jar files in a different directory on unix
box.  Loaded stylesheet on unix box.  Loaded InputFile on unix box.

saxon br saxonjarclasspath/saxon.jar -m javaclasspath -s:Input_File
-xsl:Stylesheet_File  bo Output_File

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