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