---------------------------------------------------------------- BEFORE YOU POST, search the faq at <http://java.apache.org/faq/> WHEN YOU POST, include all relevant version numbers, log files, and configuration files. Don't make us guess your problem!!! ---------------------------------------------------------------- >From going through the mail archives, I remembered to turn on all the jserv logging. [06/11/1999 22:52:20:598 CST] Status: 404 Not Found [06/11/1999 22:52:20:598 CST] Servlet-Error: ClassNotFoundException: subdir When I do: myvhost.com/servlets/subdir/MS It thinks that subdir is a servlet, rather than a directory! Perhaps, it is thinking that /MS is the translated path or something... ---------------------------------------------------------------- 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] -- -------------------------------------------------------------- Please read the FAQ! <http://java.apache.org/faq/> To subscribe: [EMAIL PROTECTED] To unsubscribe: [EMAIL PROTECTED] Archives and Other: <http://java.apache.org/main/mail.html> Problems?: [EMAIL PROTECTED]
