Programmatically specify file location within ear file

Question

Hi guys,
I am deploying an ear file under weblogic 6.1 and in part of the code I need to parse an xml file however, I can't figure out how to specify the path to the file. Whatever I try I always get an exception telling me the file cannot be found. The file itself resides within WEB-INF in the applications jar file which itself resides within the ear.
I don't know how to specify the files location.
Please help!
Simon.

Answer 1

Hi Simon,
I'm not sure if this will help - but I have had similar problems and am happy to share my experiences with you and hope they help you. (my test were done on JRun 3.0 - but am sure the behavior under weblogic is similar)
I am not sure what method you are using for accessing your xml file and what the package structure of you apps is but I assume that you are either doing a
<code>
String fileName = "myFile.xml";
File f = new File(fileName);
or
URL u = this.getClass().getResource(fileName);
</code>
I found problems with both these methods when deploying as a web app. The first method places the file in
<code> {JRunPath}\servers\default </code>
directory (which it should as the spec by default creates the file in the users default directory)
and the second creates a URL that does not correctly map to the location (see my post http://forum.java.sun.com/thread.jsp?forum=136&thread=184500)
neither of these is suitable for deploying a webapp. The only way to locate the resource seems to be to use either
<code>
fileName = ServletContext.getRealPath(fileName);
or
URL url = ServletContext.getResource(fileName);
</code>
This does leave open this issue on how to pass the servlet context down to the specific class within your application that loads the xml - the uglier option is to pass the servlet context object right down to the class that loads the resource - the other option is to use some kind of a static initializer to make the servletcontext object globally available
hope this helps