Since, I assume, you are deploying your web application into a servlet
container you are not going to be able to use a relative path to get at the
file. Tapestry isn't doing the loading of the file so using any Tapestry
related path constructs is not going to work.
Your best bet is putting your file in the classpath and using the
classloader's getResource or getResourceAsStream methods to load it.
You could also try to use the ServletContext to find where it's physically
loaded on the disk
String path = request.getSession().getServletContext().getRealPath("app");
// app is the name of your servlet/filter
But there are some significant caveats:
If you are using a secure app server then you may or may not be allowed to
read the file from this location.
If you have deployed into a WAR file then this also may not help (if the WAR
isn't exploded).
So, put it in the classpath and use the classloader to find it.
Josh
On 9/1/07, Angelo Chen <[EMAIL PROTECTED]> wrote:
>
>
> Hi,
> I have this info.xml file that I need to open, first I put it under
> WEB-INF
> and use : fr = new FileReader("info.xml"); got FileNotFoundException, I
> put
> it under images directory as the template was able to get the pictures
> there, then I use FileReader("images/info.xml"), same exception. where is
> the better place to put xml files and what's the correct path to access
> them? thanks.
>
> A.C.
> --
> View this message in context:
> http://www.nabble.com/T5%3A-xml-file-location-tf4364651.html#a12440650
> Sent from the Tapestry - User mailing list archive at Nabble.com.
>
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: [EMAIL PROTECTED]
> For additional commands, e-mail: [EMAIL PROTECTED]
>
>
--
--
TheDailyTube.com. Sign up and get the best new videos on the internet
delivered fresh to your inbox.
