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How to get rid of extra cgi-bin url components running a Flask App using wsgiref CGIHandler? – CPanel

I am on a shared cpanel hosting plan that does not support wsgi apps directly. So I have to use the wsgiref CGIHandler workaround, as described here: http://flask.pocoo.org/docs/0.12/deploying/cgi/ .

It all works and produces expected results, but there is always this extra stuff in the urls: “/cgi-bin/index.cgi/” that the python app seems to be adding automatically (to match what it detects while called by the cgi handler).

For example, I would like it to be myhost.com/login/ instead of myhost.com/cgi-bin/index.cgi/login/, or myhost.com/ instead of myhost.com/cgi-bin/index.cgi/.

All those shorter versions of the links work well, because of the engine rewrite rules are in place. I have checked that. It is just a matter of finding a way to tell the flask app to get rid of “/cgi-bin/index.cgi/”.

Some of my code:

cat www/.htaccess

# Redirect everything to CGI WSGI handler, but Don't interfere with static files
RewriteEngine On
RewriteCond %{REQUEST_FILENAME} !-f
RewriteRule ^(.*)$ /cgi-bin/index.cgi/$1 [L]

.

cat www/cgi-bin/index.cgi

#!/home/myhost/myhost.com/flasky/venv/bin/python
import os
import sys
sys.path.insert(0, '/home/myhost/myhost.com/flasky/venv/lib/python2.7/site-packages')
sys.path.insert(0, '/home/myhost/myhost.com/flasky')

from wsgiref.handlers import CGIHandler
from manage import app

CGIHandler().run(app)

.

cat www/flasky/manage.py

#!/usr/bin/env python
import os
from app import create_app, db
from app.models import User, Role, Permission
from flask_script import Manager, Shell
from flask_migrate import Migrate, MigrateCommand

app = create_app(os.getenv('FLASK_CONFIG') or 'default')
manager = Manager(app)
migrate = Migrate(app, db)

def make_shell_context():
    return dict(app=app, db=db, User=User, Role=Role, 
Permission=Permission)
manager.add_command("shell", Shell(make_context=make_shell_context))
manager.add_command('db', MigrateCommand)

if __name__ == '__main__':
    manager.run()

.

Any ideas?

Thanks!

Tags:

2

Answers


  1. Chosen as BEST ANSWER
    0

    I have found a hack for it.. but it's just a hack :-( If I override the value of SCRIPT_NAME environment variable in the wsgiref CGIHandler script, it works perfectly then.

    Here's the updated code:

    cat www/cgi-bin/index.cgi

#!/home/myhost/myhost.com/flasky/venv/bin/python
import os
import sys
sys.path.insert(0, '/home/myhost/myhost.com/flasky/venv/lib/python2.7/site-packages')
sys.path.insert(0, '/home/myhost/myhost.com/flasky')

from wsgiref.handlers import CGIHandler
from manage import app

os.environ['SCRIPT_NAME'] = ''

CGIHandler().run(app)

    .

    Basically, whatever the os.environ['SCRIPT_NAME'] value is, it gets prefixed to the flask app url every time a page is requested.

    I am still looking for a more elegant "pythonic" solution though ..


  2. 0 
    cat www/cgi-bin/index.cgi

#!/home/myhost/myhost.com/flasky/venv/bin/python
import os
import sys
sys.path.insert(0, '/home/myhost/myhost.com/flasky/venv/lib/python2.7/site-packages')
sys.path.insert(0, '/home/myhost/myhost.com/flasky')

from wsgiref.handlers import CGIHandler
from manage import app

class ScriptNameStripper(object):
   def __init__(self, app):
       self.app = app

   def __call__(self, environ, start_response):
       environ['SCRIPT_NAME'] = ''
       return self.app(environ, start_response)

    app = ScriptNameStripper(app)

CGIHandler().run(app)
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