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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