Howdy,
How about giving it a key yourself in your web.xml?  (As a context-param
or init-param to some servlet)?  That seems simpler.

Alternatively, consider using javax.servlet.context.tempdir, although
I'm not sure if that's in the SRV v2.2.

Yoav Shapira
Millennium ChemInformatics


>-----Original Message-----
>From: Jacob Kjome [mailto:[EMAIL PROTECTED]]
>Sent: Wednesday, June 12, 2002 3:48 PM
>To: Tomcat Users List
>Subject: query unique identifier of a webapp from init()?...
>
>
>I'm looking for a way to set unique system properties for any
>individual webapp.  Maybe an example of what I need will help to
>explain:
>
>For instance,  I want to set a [unique_webapp_key].log.home system
>property.
>
>so, I would do:
>System.setProperty("[unique_webapp_key].log.home")
>
>I don't want to just set "log.home" because I'm thinking that
>some other process  might override such a common name and log.home
>would end up pointing to a path that I didn't expect.
>
>I was going to use getServletContext().getServletContextName() and
>replace any spaces with ".".  However getServletContextName() was
>introduced in Servlet
>2.3 and I want it to work with Servlet 2.2.  Besides, if the
><display-name> element in the web.xml is not specified, all I get back
>is null.
>
>I also need to be able to predict the result of the unique_webapp_key
>so that I can reference it from config files.
>
>Is there something in the servlet spec 2.2+ that can help me...or am I
>overlooking something more simple?
>
>Jake
>
>
>
>--
>Best regards,
> Jacob                          mailto:[EMAIL PROTECTED]
>
>
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