Setting Python Path in Windows XAMPP using WSGI

I'm setting up a development version of a live server on Webfaction, running Django apps in a virtual Apache server environment (running without any errors) on my local machine - XP, running XAMPP Lite with Python 2.6 - which I can commit changes from via Git.

XAMPP is up and running OK with Python, and the server starts perfectly with WSGI module loaded. The problem is when I set my Python paths, they are s开发者_Go百科et half in 'nix format (with /), and half in Windows (with backslashes).

Here's the local machine Apache error, showing the corrupted python paths:

[Fri Oct 08 14:52:53 2010] [error] [client 127.0.0.1] mod_wsgi (pid=1436): Exception occurred processing WSGI script 'C:/SERVER/Python26/Lib/site-packages/website-cms/webapps/django/dev.wsgi'.
[Fri Oct 08 14:52:53 2010] [error] [client 127.0.0.1] Traceback (most recent call last):
[Fri Oct 08 14:52:53 2010] [error] [client 127.0.0.1]   File "C:/SERVER/Python26/Lib/site-packages/website-cms/webapps/django/lib/python2.5\\django\\core\\handlers\\wsgi.py", line 230, in __call__
[Fri Oct 08 14:52:53 2010] [error] [client 127.0.0.1]     self.load_middleware()
[Fri Oct 08 14:52:53 2010] [error] [client 127.0.0.1]   File "C:/SERVER/Python26/Lib/site-packages/website-cms/webapps/django/lib/python2.5\\django\\core\\handlers\\base.py", line 42, in load_middleware
[Fri Oct 08 14:52:53 2010] [error] [client 127.0.0.1]     raise exceptions.ImproperlyConfigured('Error importing middleware %s: "%s"' % (mw_module, e))
[Fri Oct 08 14:52:53 2010] [error] [client 127.0.0.1] ImproperlyConfigured: Error importing middleware cms.middleware.multilingual: "No module named cms.middleware.multilingual"

And the offending .wsgi file contents:

import os, sys

sys.path.append('C:/SERVER/Python26/')
sys.path.append('C:/SERVER/Python26/Lib/site-packages/website-cms/webapps/django')
sys.path.append('C:/SERVER/Python26/Lib/site-packages/website-cms/webapps/django/lib/python2.5')

from django.core.handlers.wsgi import WSGIHandler

#Add the path to Django itself
os.environ['DJANGO_SETTINGS_MODULE'] = 'website.settings'
application = WSGIHandler()

The Apache httpd.conf is the default for XAMPP (and not a virtual instance), with the following added to load the wsgi module

LoadModule wsgi_module modules/mod_wsgi-win32-ap22py26-3.3.so

& to point to the wsgi file:

WSGIScriptAlias / C:/SERVER/Python26/Lib/site-packages/website-cms/webapps/django/dev.wsgi

I know the XAMPP server is using Python2.6 (I'm forced to to use TortoiseGIT) and the production is on 2.5 (enfordced by the web host), but that doesn't seem to be the culprit - I would still expect to be able to set the correct path at least!

All suggestions on getting the Python path to play ball welcomed!

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My PC has Python 2.6, so I'll use all the configuration under the assumption that Python 2.6 is the target version.

1. Download newest xampp (http://www.apachefriends.org/en/xampp-windows.html), as of Nov 29, 2010, ver 1.7.3 is the newest.
2. Install xampp for windows, I installed it c:\xampp
3. Download and install Python 2.6 (http://www.python.org/download/releases/2.6/)
4. Download wsgi for windows - http://code.google.com/p/modwsgi/wiki/DownloadTheSoftware?tm=2 Refer to the documentation if necessary - http://code.google.com/p/modwsgi/wiki/InstallationOnWindows
5. Copy the so file to module directory C:\xampp\apache\modules, don't forget to rename it mod_wsgi.so
6. Add the following line to C:\xampp\apache\conf\httpd.conf
   • LoadModule wsgi_module modules/mod_wsgi.so
7. Relaunch apache using C:\xampp\xampp-control.exe

For testing I did the following steps.

1. Make C:\xampp\htdocs\wsgi\scripts directory, and copy the test test.wsgi.

The test.wsgi is as follows.

#!/usr/bin/env python
"""
A simple WSGI test application.

Its main purpose is to show that WSGI support works (meaning that the
web server and the WSGI adaptor / support module are configured correctly).

As a nice plus, it outputs some interesting system / WSGI values as a nice
HTML table.

The main use of this script will be using the WSGI "application" defined
below within your production WSGI environment. You will use some code similar
to what you see at the end of this script to use the application from that
environment. For the special case of apache2/mod_wsgi, it shoud be possible
to directly use this file.

If you start this script from the commandline either with python2.5 or with
and older python + wsgiref module installed, it will serve the content on
http://localhost:8000/ - this is mainly for debugging THIS script.

@copyright: 2008 by MoinMoin:ThomasWaldmann
@license: Python License, see LICENSE.Python for details.
"""
import os.path
import os
import sys

try:
    __file__
except NameError:
    __file__ = '?'

html_template = """\
<html>
<head>
 <title>WSGI Test Script</title>
</head>
<body>
 <h1>WSGI test script is working!</h1>
 <table border=1>
  <tr><th colspan=2>1. System Information</th></tr>
  <tr><td>Python</td><td>%(python_version)s</td></tr>
  <tr><td>Python Path</td><td>%(python_path)s</td></tr>
  <tr><td>Platform</td><td>%(platform)s</td></tr>
  <tr><td>Absolute path of this script</td><td>%(abs_path)s</td></tr>
  <tr><td>Filename</td><td>%(filename)s</td></tr>
  <tr><th colspan=2>2. WSGI Environment</th></tr>
%(wsgi_env)s
 </table>
</body>
</html>
"""
row_template = "  <tr><td>%s</td><td>%r</td></tr>"

def application(environ, start_response):
    mysite = '/Users/smcho/Desktop/django'
    if mysite not in sys.path:
        sys.path.insert(0,'/Users/smcho/Desktop/django')
    mysite = '/Users/smcho/Desktop/django/mysite'
    if mysite not in sys.path:
        sys.path.insert(0,'/Users/smcho/Desktop/django/mysite')

    """ The WSGI test application """
    # emit status / headers
    status = "200 OK"
    headers = [('Content-Type', 'text/html'), ]
    start_response(status, headers)

    # assemble and return content
    content = html_template % {
        'python_version': sys.version,
        'platform': sys.platform,
        'abs_path': os.path.abspath('.'),
        'filename': __file__,
        'python_path': repr(sys.path),
        'wsgi_env': '\n'.join([row_template % item for item in environ.items()]),
    }
    return [content]

if __name__ == '__main__':
    # this runs when script is started directly from commandline
    try:
        # create a simple WSGI server and run the application
        from wsgiref import simple_server
        print "Running test application - point your browser at http://localhost:8000/ ..."
        httpd = simple_server.WSGIServer(('', 8000), simple_server.WSGIRequestHandler)
        httpd.set_app(application)
        httpd.serve_forever()
    except ImportError:
        # wsgiref not installed, just output html to stdout
        for content in application({}, lambda status, headers: None):
            print content

2. Make C:\xampp\apache\conf\other\wsgi.conf that has the following content

This is the code

<Directory "C:/xampp/htdocs/wsgi/scripts">
  Options ExecCGI Indexes
  AddHandler cgi-script .cgi
  AddHandler wsgi-script .wsgi  
  Order allow,deny
  Allow from all
</Directory>
Alias /wsgi/ "C:/xampp/htdocs/wsgi/scripts/"
<IfModule wsgi_module>
  WSGIScriptAlias /test "C:/xampp/htdocs/wsgi/scripts/test.wsgi"
</IfModule>

3. Add this line to the httpd.conf Include "conf/other/wsgi.conf"
4. Relaunch apache.
5. You'll see wsgi info when you enter 'localhost/test' or 'localost/wsgi/test.wsgi' in your web browser.

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I also had the "server error", and went to see in the Apache error.log file: it was due to some white space or line break coming with the comment line "Its main purpose is to ..."