Can anyone give me some advice

Hi, there,

Can anyone give me some idea or gudeiance on the message "experiment for
setting up another port" posted on 29/6/03 and the latest one I posted
was on July 6, 2003. Or did I not make myself clear?

Regards
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On Tue, 08 Jul 2003 16:54:07 +0200, The Other Guy responded to a post
from "bearbear" <dahung.news.invalid@web2news.net> who wrote in
comp.infosystems.www.servers.ms-windows:

>Hi, there,
>
>Can anyone give me some idea or gudeiance on the message "experiment for
>setting up another port" posted on 29/6/03 and the latest one I posted
>was on July 6, 2003. Or did I not make myself clear?
>
>Regards


From that OP on 29/6/03:
<quote>
C:>apache -f C:>pogram files/apache group/apache2/conf/httpd8080.conf

and hit the <enter> key so as to start another server, but the
computer
responds me that 'apache' is not an internal or external command.

Can anybody help me to solve this problem? what is going on?
</quote>

You need to do one of two things:
i) in DOS, navigate (i.e., cd) to the C:>\program files\apache
group\apache2\bin directory to execute the "apache -f" command; or
ii) to execute the apache command from anywhere in the command line
prompt location, add into the autoexec.bat the path to the apache
binary executable (i.e., C:>\program files\apache group\apache2\bin)
so that it is read at startup.


I note in the original path that you have:
C:>pogram files/apache group/apache2/conf/httpd8080.conf
^^^

I am assuming that this was a typo here and that you meant to
have"program files"; but if you typed what you said into the command
line you would get an error because it is likely the wrong path.

From your description in your post of July 6, 2003, there was no
problem. Apache appeared to be already running and you were trying to
start it when it was already started (see Jesse's reply).

HTH
TOG

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