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> on 11/3/99 11:01 AM, Jim Heintz <[EMAIL PROTECTED]> wrote:
>
> > Does anyone know what the logical or physical limit to the number of
> > zones one can create within a JServ configuration is?
>
> It should be in the thousands.
Is there a performance/memory penalty incurred when you use a lot of zones?
Ie: When I add a zone, how much slower is my jvm and how much more memory
am I consuming?
This is becoming a concern for us, since we are unable to use subdirectories
in our zones*.
(*I have read the faq about this, and it doesn't work. Has anyone been
successful doing that?)
--
Chris
my original post about zone subdirs:
We run all of our servlets in servlet zones, one per virtual host.
So on one of our virtual hosts we mapped https://vhost/servlets to
.../zones/vhost (We put all of our zones in a zones dir.)
Well, I put MyServlet.java in .../zones/vhost/test/ and changed the package
to "package test;" and recompiled. But, it gives me "HTTP 404 - File not
found."
Now from the FAQ it says:
"It looks in the 'test' subdirectory of each directory in the classpath for
files in the 'test' package."
Well, for the heck of it, I added .../zones/vhost to the classpath of
jserv.properties. Well that didn't fix it. Which even if it did, wouldn't
be right, since I don't want one zone to accidently import classes from
another zone.
Any suggestions?
--
Chris Busch [EMAIL PROTECTED]
MedServe Link Inc.
>From the FAQ:
I have a directory within my servlets directory and things break when I try
to use http://www.server.com/servlets/test/TestServlet, but when I put
TestServlet directly in the servlets directory and remove the "test" from
the URL, things work just fine. Is this a bug? No, this is not a bug but a
wanted feature. This problem derives from the way Java structures classes
into subdirectories per package name. People suffer similar problems writing
non-servlet code. (we have a love-hate relationship with this approach: it's
clean, but it creates deep directory trees.) Java would expect a file called
$DIR/test/SimpleServlet to have the fully-qualified classname
test.SimpleServlet. It looks in the 'test' subdirectory of each directory in
the classpath for files in the 'test' package.
The module currently converts a request for test/SimpleServlet into a
request for the class test.SimpleServlet. It finds the class file, but gets
confused because the file actually contains a definition of the file
'SimpleServlet'. In short: make sure the package name declared in the class
matches the directory structure where the class file is located.
To fix this particular problem: declare the servlet using a proper package
name: org.dummy.test.SimpleServlet, compile it into
$ROOT/servlets/org/dummy/test/SimpleServlet.class, and make your ServletPath
$ROOT/servlets. Add an alias or servlet.simple.code property to generate a
shorter URL if you like. (Thanks to Martin Pool)
--
Chris Busch [EMAIL PROTECTED]
MedServe Link Inc.
--
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