test message - sorry
-Original Message-
From: Tim Moore [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, December 17, 2002 1:11 PM
To: Tomcat Users List
Subject: RE: File access from a servlet.
-Original Message-
From: Patrick Martz [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, December 17, 2002 2:02 PM
To: 'Tomcat Users List'
Subject: RE: File access from a servlet.
Ok well that's exactly the problem. getResourceAsStream
requires you to supply the path of the resource and that is
what I'm missing. I did a quick look at ServletContext and
iterated through the attributes and found none that seemed to
give me what I want. These are the attributes currently
defined:
org.apache.catalina.jsp_classpath
javax.servlet.context.tempdir
org.apache.catalina.resources
org.apache.catalina.WELCOME_FILES
None of which seems to be what I'm looking for...essentially
something that will tell me the path of my current context so
I can modify that path to access my data file. :)
But the path is relative to the root of the context, so you don't need
to know the path to the context, just where the file is within it. For
example, if your file is actually in
/usr/local/tomcat/webapps/myapp/data/blah.dta, you would call
ServletContext.getResourceAsStream(/data/blah.dta).
If you *really* want the path of the context root, you can use
ServletContext.getRealPath(/) (or
ServletContext.getRealPath(/data/blah.dta) to get the file's path) but
that won't work with WARs, and shouldn't really be necessary.
I just reread your message and realized that you said that the data is
stored in the directory with the servlet class file. Maybe you can use
getClass().getResourceAsStream(blah.dta)?
--
Tim Moore / Blackboard Inc. / Software Engineer
1899 L Street, NW / 5th Floor / Washington, DC 20036
Phone 202-463-4860 ext. 258 / Fax 202-463-4863
-Original Message-
From: Tim Moore [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, December 17, 2002 10:46 AM
To: Tomcat Users List
Subject: RE: File access from a servlet.
-Original Message-
From: Patrick Martz [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, December 17, 2002 1:42 PM
To: '[EMAIL PROTECTED]'
Subject: File access from a servlet.
Hi all.
I'm currently working on a java servlet with tomcat and I
want it to be able to load a different data file dependent on
certain parameters passed to the servlet. The problem is that
if I just try to open the file with the file name (i.e.
FileInputStream fin = new FileInputStream(blah.dta);) it
fails to find the file. I am guessing this is because the
runtime directory is different from the directory the servlet
is running in? (the data file and the servlet are in the same
directory, but the servlet fails to find the file still). So
my question is, is there a way to get the current runtime
directory for Tomcat so that I can perhaps supply a relative
path to get to the file and have the servlet be able to open
it? Thanks!
Patrick
P.S. For debugging purposes I HAVE tested opening of the file
from a stub class and it works just fine that way, but fails
from the servlet.
Rather than using FileInputStream, try
ServletContext.getResourceAsStream. This is the preferred method for
accessing files within your webapp. You pass in the path relative to
the context root directory. An added bonus is that this will
still work
if you deploy your webapp as a WAR.
http://java.sun.com/j2ee/sdk_1.3/techdocs/api/javax/servlet/ServletConte
xt.html#getResourceAsStream(java.lang.String)
--
Tim Moore / Blackboard Inc. / Software Engineer
1899 L Street, NW / 5th Floor / Washington, DC 20036
Phone 202-463-4860 ext. 258 / Fax 202-463-4863
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