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


Subject: [xsl] How do I properly define a unix saxon script command line
From: "Catherine Wilbur cwilbur@xxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Tue, 11 Nov 2014 21:46:54 -0000

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

_____________________________________________________________________
Catherine Wilbur  |  Senior Application Programmer  |  IT Services
401 Sunset Avenue, Windsor ON Canada  N9B 3P4
(T) 519.253.3000 Ext. 2745  |  (F) 519.973.7083  |  (E)
cwilbur@xxxxxxxxxxx
www.uwindsor.ca/its


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