Apache mod_wsgi error: Forbidden You don't have permission to access / on this server

Question

I'm using Ubuntu 10.04.
I create a django project under /home/wong2/Code/python/django2/ named atest
and create a wsgi file setting.wsgi in the same directory
Here is the content of setting.wsgi :

import os 
import sys

path = '/home/wong2/Code/python/django2'

if path not in sys.path:
    sys.path.append(path)
os.environ["DJANGO_SETTINGS_MODULE"] = "atest.settings" 
from django.core.handlers.wsgi import WSGIHandler 
application = WSGIHandler()

and here is what I add to my my httpd.conf:

<VirtualHost *:80>
    ServerName localhost
    WSGIScriptAlias / /home/wong2/Code/python/django2/setting.wsgi
    <Directory />
        Options FollowSymLinks
        AllowOverride None
        Order deny,allow  
        Allow from all 
    </Directory>
    <Directory "/home/wong2/Code/python/django2/atest">
        Order allow,deny
        Allow from all
    </Directory>
</VirtualHost>

The problem is, when I visit http://localhost, it says

Forbidden

You don't have permission to access / on this server.

Thanks a lot.

Answer 1

The second directory block doesn't match where you have your WSGI script file installed. It is very bad practice though to stick the WSGI script file in a location where source code or other sensitive files exist, ie., same directory or sub directory. Instead you should stick it in a sub directory of its own. Thus:

WSGIScriptAlias / /home/wong2/Code/python/django2/atest/apache/setting.wsgi
<Directory "/home/wong2/Code/python/django2/atest/apache">
    Order allow,deny
    Allow from all
</Directory>

So, create 'apache' subdirectory under 'atest'. Move 'setting.wsgi' into that 'apache' subdirectory and change config to above.

Your problem also may be caused by restrictive permisions on your home directory as Apache cannot see inside.

Go watch:

http://code.google.com/p/modwsgi/wiki/WhereToGetHelp?tm=6#Conference_Presentations

as it explains these permissions problems as well as issues like where to stick your code and the WSGI script file.

Also read:

http://code.google.com/p/modwsgi/wiki/IntegrationWithDjango

Answer 2

With Django 1.5+ you should use the suggested way described in the documentation:

WSGIScriptAlias / /path/to/mysite.com/mysite/wsgi.py
WSGIPythonPath /path/to/mysite.com

<Directory /path/to/mysite.com/mysite>
<Files wsgi.py>
Require all granted
</Files>
</Directory>

https://docs.djangoproject.com/en/1.7/howto/deployment/wsgi/modwsgi/

Answer 3

I had the same problem due to restricted file permissions.

By default your home folder user has the settings:

ls -l /home/

drwx------ 36 user user 12288 Nov 28 20:56 user

meaning that no one else but yourself is able to even browse that folder.

Adding execute option to the folder fixed my problem

chmod o+x /home/user/
ls -l /home/

drwx-----x 36 user user 12288 Nov 28 20:56 user

Answer 4

try http://localhost/index.html does that show the apache page?

If you wsgi is configured on root correctly it wont.

You get that error message because you might be asking the webserver for the directory listing of www folder

edit:

My projects are always in

/var/pyproj//pysrc/ and server files i put in /var/py_proj//server/

Here is my go to wsgi script

import os, sys

import logging
logger =logging.getLogger("")
logger.setLevel(logging.DEBUG)
handler = logging.StreamHandler(sys.stderr)
handler.setLevel(logging.DEBUG)
formatter = logging.Formatter('%(levelname)-8s %(messages)s')
handler.setFormatter(formatter)
logger.addHandler(handler)

sys.stdout = sys.stderr
from os.path import abspath, dirname, join
import site

PROJECT_ROOT = abspath(join(dirname(__file__), "../pysrc"))
SETTINGS_LOC = abspath(join(PROJECT_ROOT, ''))
site.addsitedir(SETTINGS_LOC)
os.environ["DJANGO_SETTINGS_MODULE"] = "settings"

from django.core.management import setup_environ
import settings
setup_environ(settings)

sys.path.insert(0, join(PROJECT_ROOT, "apps"))

from django.core.handlers.wsgi import WSGIHandler

application = WSGIHandler()

and i use a vhost.conf

WSGIDaemonProcess pinax_demo user=www-data group=www-data threads=10 python-path=/var/.virtualenvs/pinax_demo/lib/python2.6/site-packages

<Directory /var/py_proj/pinax_demo/server/>
    Order deny,allow
    Allow from all
</Directory>

Alias /pinax_demo /var/py_proj/pinax_demo/server/pinax_demo.wsgi
<Location /pinax_demo/>
    #WSGIProcessGroup must match the WSGIDaemonProcess
    WSGIProcessGroup pinax_demo
    SetHandler wsgi-script
    Options +ExecCGI
</Location>

which i include in the sites-available folder in a file that looks like this

<VirtualHost *:8080>
        DocumentRoot /var/www/

        <Directory /var/www/>
                Order deny,allow
                Allow from all
                Options -Indexes FollowSymLinks
                DirectoryIndex index.php
                AllowOverride None
        </Directory>

        ServerAdmin oto@example.com

        LogLevel error
        ErrorLog  /var/srv_logs/apache_error.log
        CustomLog  /var/srv_logs/apache_access.log combined
        #insert
        Include /var/py_proj/pinax_demo/server/vhost.conf
        Include /var/py_proj/<other>/server/vhost.conf
        Include /var/py_proj/<other>/server/vhost.conf

</VirtualHost>

httpd.conf is empty. Apache is also terrible at sending static files so i have my apache setup on port 8080 and nginx setup on port 80 which forwards to apache. You should be able to just change the port on my sites-avaiable file to run it without nginx but i would not go live that way

Answer 5

You may get a surprise where your WSGI has its home directory! On my admittedly naive wsgi_mod installation, it is / - I was so surprised I had to write this script to make sure:

import os

def application(environ, start_response):
    status = '200 OK'; e=os.listdir(os.getcwd())
    output = '''WSGI %s<hr>'''%(e)
    response_headers = [('Content-type', 'text/html'),
                        ('Content-Length', str(len(output)))]
    start_response(status, response_headers)
    return [output]

That means imported applications will need a full path to be found, e.g.:

sys.path.append('/home/foo/www/Forms')
from DataCommon import page

This is the recommended (by Google) approach to this:

import sys; path='/home/foo/www/Forms';
if path not in sys.path: sys.path.append(path)