Dear All, i am new to web2py and this forum, so please pardon me if this subject has already been addressed. Here's the problem:
i have python 2.7 installed on windows 7. i have downloaded web2py for windows from this site: http://web2py.com/examples/default/download. According to the site: "The source code version (of web2py) works on all supported platforms, including Linux, but it requires Python 2.4, 2.5, 2.6, or 2.7. It runs on Windows and most Unix systems..." After downloading the file, ii am then instructed to "...unzip it and click on web2py.exe (windows)..." After clicking on web2py.exe, i get the following message on the command line: "web2py Web Framework Created by Massimo Di Pierro, Copyright 2007-2011 Version 1.98.2 (2011-08-04 00:47:09) Database drivers available: SQLite3. pymysql Starting hardcron..." Another window pops up with the following information: Server IP: 127.0.0.1 Server Port: 8000 Choose Password: When i enter a password and either press enter or click the "start server" button, i get the following command line error message: ERROR: Rocket.Errors. Port8000: Socket 127.0.0.1:8000 in use by other process and it won't share. Please visit: http://127.0.0.1:8000 starting browser... WARNING: Rocket.Errors.Port8000: Listener started when not ready. Can anyone help a newbie? Otherwise it seems my adventure with web2py ended before it even started!
