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Posted to dev@tomcat.apache.org by bl...@lgenc.co.kr on 2000/02/22 02:12:48 UTC

newbie question about port number..

Hi...

Sorry for this newbie question...

When I access Servlet :8080 port number should be added to locahost...

But I would like to not type the port number...

How can I access Servlet without port number?...

Thanks in advance..




How about thread???

Posted by airof98 <ai...@koreansource.com>.
> shin hee su in south korea write..

I installed tomcat.
test it .      it was very well..
jsp is also.
So I tried to run my servlet.
it was multi-thread servlet.
like this.
       Thread a = new  mythread(out);
       a.start();
       Thread b = new mythread(out);
       b.start();
out is for printing content. and it was synronized.

first i tested it with JSDK --servletrunner . it go well.
and JRUN is also..
but tomcat takes a lot of time and didn't end...
what is problem...

above program for search-engin which search shopping mall and auction site.
thanks.




Tomcat/JServ/SSL

Posted by Mike Engelhart <me...@earthtrip.com>.
I just went through the FAQ, the archives at mail-archive.com and the tomcat
archives about this and saw some threads with the same topic, but no
answers.  

here's what I have
RedHat 6.1
Apache 1.3.11
Mod_ssl 2.5.0 (compiled against OpenSSL 0.9.4)
JServ 1.1 Final
Tomcat 3.1M1 (also tried Tomcat 3.0 final)

When using Tomcat as the servlet engine behind JServ, headers don't get
passed through JServ.  Even if I start the server using ./apachectl start
instead of ./apachectl startssl, I get no header information sent through to
my servlets.   For example the RequestHeaders example servlet just returns a
blank page.  User-Agent, Accept-Language,etc.  None of these are passed
through.

I've tried recompiling everything as static and everything as DSO and
combinations thereof and nothing seems to work.
The one thing that DOES work with all of this is if I replace JServ 1.1
Final with JServ 1.1b3.  everything works fine with that release on both
Tomcat 3.0 and Tomcat 3.1M1.   I don't know how many changes occured in
between the b3 and final release but somewhere something either broke or was
fixed and broke something in Tomcat.

Any help would be appreciated.  I posted to the JServ list which was no
help.

Thanks,
Mike


Re: newbie question about port number..

Posted by Daniel Kehoe <ke...@fortuity.com>.
>When I access Servlet :8080 port number should be added to locahost...
>But I would like to not type the port number...
>How can I access Servlet without port number?...

By default, Tomcat is accessible on port 8080 (presumably so it 
doesn't interfere with any previously installed web servers). Web 
browsers connect to port 80 by default, so you should change the 
Tomcat configuration file
<your directories>/tomcat/server.xml
substituting
<Server adminPort="-1" workDir="work">
     <ContextManager port="8080" hostName="" inet="">
with
<Server adminPort="8080" workDir="work">
     <ContextManager port="80" hostName="" inet="">
and restarting Tomcat. If you are not starting Tomcat as the system 
superuser (root), you will get an error "java.net.BindException: 
Permission denied" because access to port 80 is denied to anyone but 
root. With Tomcat set to use port 80, you must be the superuser to 
start and stop Tomcat.

With Tomcat listening on port 80, you should be able to use a web 
browser to access Tomcat test pages at
http://www.yourserver.com/examples/


RE: newbie question about port number..

Posted by Angsuman <an...@technologist.com>.
Configure tomcat to run on port 80, which is the default port.
Then you can access the servlets without explicitly typing the port number.

-----Original Message-----
From: blueless@lgenc.co.kr [mailto:blueless@lgenc.co.kr]
Sent: Monday, February 21, 2000 5:13 PM
To: tomcat-dev@jakarta.apache.org
Subject: newbie question about port number.. 


Hi...

Sorry for this newbie question...

When I access Servlet :8080 port number should be added to locahost...

But I would like to not type the port number...

How can I access Servlet without port number?...

Thanks in advance..




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