Starting Apache 2.2 on Windows (applies to *nix also)

I recently downloaded XAMPP the other day, and had a little trouble starting the inbuilt Apache 2.2 web server. Here is what i did to resolve these issues.
(This also applies to NIX systems).

I attempted to start the the apache service [apache_installservice.bat], but I received a strange error. So i decided to open the Apache Monitor manually and start the webserver.

Monitor Location: .\apache\bin\ApacheMonitor.exe

I executed the start button and the server failed to start with an ambiguous error message. I opened the event log and found the following error...

The Apache service named reported the following error:
>>> (OS 10048)Only one usage of each socket address (protocol/network address/port) is normally permitted. : make_sock: could not bind to address 0.0.0.0:80

It seemed i already had programs/services running on port 80, so i opened up the Apache config file and altered the port.

Config Location: .\apache\conf\httpd.conf

Change: Listen 80 ---> to: Listen 8080

This will run the server on 8080 rather than 80. (assuming 8080 is free!)... you can check "netstat" on "Start>Run>cmd" to check this.



I then attempted to fire up the web server again, but the same error occured (but with a different port!) This time the port being 443...

The Apache service named reported the following error:
>>> (OS 10048)Only one usage of each socket address (protocol/network address/port) is normally permitted. : make_sock: could not bind to address 0.0.0.0:443 .

This port tells Apache which port SSL associated with. This can also be modified using the SSL config file.

SSL Config file location: .\apache\conf\extra\httpd-ssl.conf

Change: Listen 443 ---> to: Listen 444

(Providing nothing is running on 444)


I then attemted to start the Apache Server, and everything ran smoothly!