I still have questions about the path to a file. {This is within a
servlet... actually in my case I am trying to read/write from a
ContextListenr} I understand that just opening a FIle will be relative
to where Tomcat was started. Is there a simple java method (and on what
class) that will give me the correct path to 'webapps/MyApplication/'.
As recommended below, you could use Class.getResourcesStream(), but as
I understand it, that will give the location of this class under
WEB-INF. And that assumes you are calling it from one of your own
classes and the loader picked up your class.
I got the following technique from a book, but I am not in love with it:
public void contextInitialized(ServletContextEvent event) {
ServletContext sctx = event.getServletContext();
String propPath = sctx.getRealPath( "/WEB-INF/resource.properties");
FileInputStream inStrm = new FileInputStream(propPath);
.....
}
Is this the proper way to do it?
-d
Caldarale, Charles R wrote:
From: popprem [mailto:popp...@gmail.com]
Subject: Path problem
David answered the critical part of your question, but I thought I'd take a
crack at the rest.
That means tomcat defaultly points to bin directory
"Tomcat" doesn't defaultly [sic] point to anything. The current directory of
the process you used to start Tomcat when using the script happened to be Tomcat's bin
directory - but that's just the way you're using the script, not a Tomcat requirement.
When i used tomcat.exe & installed tomcat in my machine & run
the program,i found that the test file i created in my code
level was in windows/system32/ directory.
Because that's the default current directory for services.
It's almost always a bad idea to depend on the current directory setting in a
complex server environment; if you must have a file system path to something,
better to use one based off a system property. In your case, the
getResourceAsStream() method is much preferred, since it elinates any file
system dependencies.
- Chuck
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