Unfortunately, there is no means within the init() method of a servlet to
determine the context path of this web application (i.e. the
'/APPLE" part of a complete request URI). You won't know that until the
first request comes in.
Craig McClanahan
On Mon, 6 Aug 2001, Diu Lee Lo Mo wrote:
>
> Dear all,
>
> I would like to ask how to get the web virtual path
> of a specified project in tomcat.
>
> E.g. I got a project APPLE which are placed all
> source in /webapp/APPLE.
>
> By default, this project can be accessed by
> Http://localhost:8080/APPLE/
>
> Assume I am running a servlet called 'EAT.class' ,
> then the path should be something like that
> Http://localhost:8080/APPLE/servlet/EAT .
>
> How can EAT.class know this path when running INIT()
> function ?
>
> I know that we can get some parameters from web.xml
> by using getInitParameter() function. I also know that
> we can get the virtual path by using
> request.getRequestURI() inside doPost()/doGet()
> function. But I don't know how to get it inside INIT()
> function.
>
> Thank you for your help!
>
> M.T.
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