Thanks for assistance your files doesn't work for me.   But, i've some half solution: i just used 2 htaccess, one in public_html folder with this code:   RewriteEngine On RewriteBase / RewriteRule ^(media/.*)$ - [L] RewriteRule ^(admin_media/.*)$ - [L] RewriteRule ^(dispatch\.wsgi/.*)$ - [L] RewriteRule ^(.*)$ /dispatch.wsgi/$1 [QSA,PT,L]   and the other in my project folder:     RewriteEngine On RewriteBase / RewriteRule ^(media/.*)$ - [L] RewriteRule ^(admin_media/.*)$ - [L] RewriteRule ^(dispatch\.wsgi/.*)$ - [L] RewriteRule ^(.*)$ expert2/dispatch.wsgi/$1 [QSA,PT,L]   and dispatch.wsgi file like this;     import os, sys     sys.path.append("/home1/medreda/public_html/expert2")   ### EDIT LINE ABOVE TO YOUR SITE ###   os.environ['DJANGO_SETTINGS_MODULE'] = 'settings' os.environ['PYTHON_EGG_CACHE'] = '/home1/medreda/.python_egg_cache'   ### YOU NEED TO CREATE THE DIRECTORY ABOVE ###     def application(environ, start_response):   """Simplest possible application object"""   output = "Hello World from hosted WSGI"   status = '200 OK'   response_headers = [('Content-type', 'text/plain'), ('Content-Length', str(len(output)))]   start_response(status, response_headers)   return [output]   It gives me : Hello word in my browser with this URL pydevstore.co.cc (for information it doesn't call any web app here).   But when i use this code for the wsgi file (given in the example):   import os, sys     sys.path.append("/home1/medreda/public_html/expert2")   ### EDIT LINE ABOVE TO YOUR SITE ###   os.environ['DJANGO_SETTINGS_MODULE'] = 'settings' os.environ['PYTHON_EGG_CACHE'] = '/home1/medreda/.python_egg_cache'   ### YOU NEED TO CREATE THE DIRECTORY ABOVE ###     import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler()   it gives an error (when i call my django app)   So could you say what does it mean, please