Deploy an Oracle JSP application
Hi there,
I created a very simple JSP application based on "Scott" and tried to deploy it on Tomcat 3.2. I generated two archives. One is ScottBC.jar which has anything in the BC4J project for "Scott". Another one is ScottClient.jar which only has connections.properties and ScottClient_ScottBC_ScottBCModule.properties.
I got the following error message when I tried to access main.jsp.
java.lang.RuntimeException: JSP Registry could not locate runtime property file: ScottClient_ScottBC_ScottBCModule.properties
Do you know where I went wrong in this case?
Thanks in advance.
Rick
null
the classpath can not recognize the *.properties file. there are two solutions:
1. put the *.properties file into a *.jar file and set *.jar file into classpath.
2. set a classpath like
wrapper.classpath=path\
where "path" is a directory. Put the *.porperties file into the directory directly.
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=============================
type Exception report
message Internal Server Error
description The server encountered an internal error (Internal Server Error) that prevented it from fulfilling this request.
exception
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====================
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nullOK, I changed my classpath jserv.properties file as you suggest, but I get the following error when I run my applet from IE:
wrapper.bin=K:\Ora816\Apache\jdk\bin\java.exe
wrapper.env=PATH=K:\Ora816\bin
# Oracle XSQL Servlet
wrapper.classpath=K:\Ora816\lib\oraclexsql.jar
# Oracle JDBC (8.1.7)
wrapper.classpath=K:\Ora816\jdbc\lib\classes12.zip
# Oracle XML Parser V2 (with XSLT Engine)
wrapper.classpath=K:\Ora816\lib\xmlparserv2.jar
# Oracle XML SQL Components for Java
wrapper.classpath=K:\Ora816\rdbms\jlib\xsu12.jar
# XSQLConfig.xml File location
wrapper.classpath=K:\Ora816\xdk\lib
# Oracle Servlet
wrapper.classpath=K:\Ora816\lib\servlet.jar
# Oracle Java Server Pages
wrapper.classpath=K:\Ora816\jsp\lib\ojsp.jar
# Oracle Util
wrapper.classpath=K:\Ora816\jsp\lib\ojsputil.jar
# Oracle Java SQL
wrapper.classpath=K:\Ora816\sqlj\lib\translator.zip
# Oracle JDBC
wrapper.classpath=K:\Ora816\jdbc\lib\classes12.zip
# SQLJ runtime
wrapper.classpath=K:\Ora816\sqlj\lib\runtime.zip
# Oracle Messaging
wrapper.classpath=K:\Ora816\rdbms\jlib\aqapi11.jar
wrapper.classpath=K:\Ora816\rdbms\jlib\jmscommon.jar
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wrapper.env=ORACLE_HOME=K:\Ora816
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wrapper.env=ORACLE_SID=ORAWEB
wrapper.env=LD_LIBRARY_PATH=K:\Ora816\lib
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wrapper.classpath=K:\Ora816\Apache\BC4J\lib\aurora_client.jar
wrapper.classpath=K:\Ora816\Apache\BC4J\lib\Bc_Hr_Main1.jar
wrapper.classpath=K:\Ora816\Apache\BC4J\lib\classes12.zip
wrapper.classpath=K:\Ora816\Apache\BC4J\lib\connectionmanager.zip
wrapper.classpath=K:\Ora816\Apache\BC4J\lib\javax_ejb.zip
wrapper.classpath=K:\Ora816\Apache\BC4J\lib\jbo8iclient.zip
wrapper.classpath=K:\Ora816\Apache\BC4J\lib\jbodatum12.zip
wrapper.classpath=K:\Ora816\Apache\BC4J\lib\jboeb.jar
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wrapper.classpath=K:\Ora816\Apache\BC4J\lib\jbomt.zip
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wrapper.classpath=K:\Ora816\Apache\BC4J\lib\vbjorb.jar
wrapper.classpath=K:\Ora816\Apache\BC4J\lib\xmlparserv2.jar
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Xml files deployment in a jsp application
I have a xml file and i want to read it in a jsp file.I use some java classes to read the xml file.I use DOM.
In the jsp file i'm calling a function from a clas that returns a vector with elements from the xml file.
Problem:If the same function is called by a main function (not in web app) it works just fine,but if the function is called in a jsp file it does not see the xml file!!!!
Question:Where do i have to place my xml file that my function called in a jsp file to see (use) the xml file??
Describe the exact location please....the directory
Please help me.I've tried everything.
Thank youFor example if you put your XML under WEB-INF then from JSP it can be accessible as:
String path = session.getServletContext().getRealPath("/WEB-INF/myfile.xml");
.. read the file code goes here ...
P.S. That will work if your application is not packed into WAR.
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