Or you could let go of the idea of file paths and use a URL from the ServletContext.getResource() or the corresponding InputStream from ServletContext.getResourceAsStream(). This also allows you to package the .xsl up in a .war file without breaking.
-----Original Message----- From: Mark Petrovic [mailto:[EMAIL PROTECTED] Sent: Mon 5/15/2006 9:07 PM To: Tomcat Users List Subject: Re: Determining file locations in tomcat You can put the path in an <init-param> in the application's web.xml. Then retrieve it at runtime in your servlet.init() method using getInitParameter("path"). On 5/15/06, Calvin Deiterich <[EMAIL PROTECTED]> wrote: > > I need to place a file(.xsl) in with a tomcat servlet application and > I do not want to hard-code the path into the java code. I want the > file to be in the \webapps\<projectname>\WebContent\xml\xslt\ path. > What is the best way to handle this. > Thanks > Calvin > > -- > http://mywebpages.comcast.net/tnczoo/ > ----- > "If a dog will not come to you after having looked you in the face, > you should go home and examine your conscience." > - Woodrow Wilson > ----- > > > > --------------------------------------------------------------------- > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > > -- Mark AE6RT --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
