Question

I want to set a path in my web.xml which is relative to the WEB-INF or the WEB-INF/classes.

However, the following code (repository.home) seems to pick up a path relative to c: root drive path on my windows machine here.

    <servlet>
  <servlet-name>ContentRepository</servlet-name>
  <servlet-class>org.apache.jackrabbit.servlet.jackrabbit.JackrabbitRepositoryServlet</servlet-class>
  <load-on-startup>1</load-on-startup>
    <init-param>
      <param-name>repository.home</param-name>
      <param-value>/WEB-INF/classes/testit</param-value>
    </init-param>
</servlet>

<servlet>
  <servlet-name>JackrabbitServlet</servlet-name>
  <servlet-class>com.orsa.seam.JackrabbitServlet</servlet-class>
  <load-on-startup>2</load-on-startup>
</servlet>

Answer 1

That parameter is OK, i.e. when you read it, it will be still a String with the value:

/WEB-INF/classes/testit

A parameter in web.xml file is just a parameter – it’s not a “path”.

So the question is what you do with that parameter?

Since you are trying to access resources(files) inside a servlet container, you need to use specific methods, e.g. something like

ServletContext#getRealPath()

to get the real path of e.g. /WEB-INF/classes/testit, than

• append to it what you need (thus having a full path to those files inside that directory), or subtract it from other “full paths”, thus having the relative paths to it. E.g. if you have ‘/testit/test1.txt’ ‘/testit/test2.txt’ you can access those files by just appending them to the real path obtained above.

• Or subtract that path from other ‘full paths’, thus having the relative path to your parameter. E.g. if you have the full path (c:/tomcat/webapps/myapp/WEB-INF/data/testit/test.txt), than you just need to subtract that full path of your parameter, so you will obtain the relative path: ‘/testit/test2.txt’