Did you tried it?
I think and i read it on this list, that this is a bug/mistake in the description and 
it should mean 2.0.43 and above.
But read the archive before you try it.

Mike

-----Ursprüngliche Nachricht-----
Von: Yeray Santana Borges [mailto:[EMAIL PROTECTED] 
Gesendet: Dienstag, 11. November 2003 13:27
An: Tomcat Users List
Betreff: RE: Apache+Tomcat+Windows, how?

ok, thanks all

My problem is that i have apache web server 2.0.48 and the mod_jk.dll is
only for apache 2.0.43. Iīm searching for other version of mod_jk.dll for
2.0.48 but i canīt find it. Other solution is download another version of
apache server that work with that dll.

Thanks for help me, at least i have could know what was my problem.

-----Mensaje original-----
De: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Enviado el: martes, 11 de noviembre de 2003 8:55
Para: Tomcat Users List
Asunto: Re: Apache+Tomcat+Windows, how?


The best how to:
www.gregoire.org/howto/Apache2_Jk2_TC4.1.x_JSDK1.4.x.html

or:
http://johnturner.com/howto/winxp-howto.html

Mike

Found in jakarta

Scrive Bill Barker <[EMAIL PROTECTED]>:

> Go to http://jakarta.apache.org/site/binindex.cgi, scroll down to "Tomcat
> WebServer Connectors", click on "JK 1.2 Binary Releases", click on
"win32",
> and, what do you know ;-).
>
> "Yeray Santana Borges" <[EMAIL PROTECTED]> wrote in message
> news:[EMAIL PROTECTED]
> > Hi all,
> >
> > Iīm a spanish newbie using Tomcat. My enviroment: Apache HTTP Server
> 2.0.48,
> > Jakarta Tomcat 4.1 and w2000 Server.
> >
> > I canīt connected the Apache Server and Tomcat. I get the binary
> > distribution and i canīt find the mod_jk.so that i must include in the
> > Apache web with LoadModule directive.
> > I get the mod_jk.dll bat after edit http.conf my server donīt start
> because
> > can find API in mod_jk.dll. I know that i must use Loadmodule
"something"
> > .... but i donīt know what.
> >
> > Anybody help me with this configuration? how can i do it?
> >
> > Thanks, sorry if my english is not correct, iīm a spanish student from
> > Canary Island.
>
>
>
>
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