wsgi

问题 I have a Python 2.7.6 Flask application that is trying to parse a SAML XML document using the lxml library. I'm running into an issue where etree.tostring(...) returns an empty string. etree_string = etree.tostring(etree.fromstring(b'<test1><test2></test2></test1>')) return etree_string # output: '' This appears to only occur when the code is run within the Flask app, served by mod_wsgi in Apache. I say this because in the same virtualenv, if I open a python interpreter and run: >>> etree

问题 I have a small web application that I have built using Flask and python. With the internal server that I used for developing everything runs fine. However now I want to use apache to start using it. But it doesn`t work. Keep in mind that I have never worked with apache or web based stuff before. I used this guide as my starting point: http://flask.pocoo.org/docs/deploying/mod_wsgi/ right now I have my application which is in the file called "/rg/server.py" and looks like this: app=Flask(_

问题 I am working on a toy web framework and I have implemented a method redirect_to for redirecting the user to a specified URL. Here the code def _redirect_to(_self, _url): """Takes an url as argument and returns a location header""" return ([('Location', _url)], "") # --- (i) The returned tuple is then handled by the app, here's the code ... else: status = '200 OK' params['_GET'] = get_data(_environment) params['_POST'] = post_data(_environment) (headers, response) = objview(objcontroller,

问题 Output from client.py is text/plain although no content-type header was sent to the server. Why? # ---------------------------------------------- server.py from wsgiref.simple_server import make_server def simple_app(environ, start_response): print environ.get('CONTENT_TYPE') start_response('200 OK', []) return [] make_server('localhost', 88, simple_app).serve_forever() # ---------------------------------------------- client.py import subprocess import urllib2 p = subprocess.Popen(['python',

问题 I want Gunicorn to talk with TileStache via WSGI. But when I run this command... gunicorn "TileStache:WSGITileServer('/var/osm/bright/project/OSMBright4/tilestache.cfg')" ...I get these errors: 2013-03-30 23:02:41 [14300] [INFO] Starting gunicorn 0.17.2 2013-03-30 23:02:41 [14300] [INFO] Listening at: http://127.0.0.1:8000 (14300) 2013-03-30 23:02:41 [14300] [INFO] Using worker: sync 2013-03-30 23:02:41 [14305] [INFO] Booting worker with pid: 14305 Error loading Tilestache config: 2013-03-30

问题 I have a Centos 7 server with cPanel and I'm working on a Telegram bot for my business needs. The bot should be able to run a terminal command with os.system or subprocess.Popen, however both options do not work when configured through a webhook + wsgi process. I tested both with bot.polling method and they worked as a charm, however after I switched to webhook method served by flask and wsgi, both stopped working for me. I have tried the following: mycommand = "python3.6 GoReport.py --id 31

问题 I try 2 way to config wsgi to use Django for my site. my root project file is in /var/run/myApp. this is my wsgi.py : import os os.environ['DJANGO_SETTINGS_MODULE'] = 'myApp.settings' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler() 1.I install everything that need to run Django on apache and using centos as OS. I use this site to run wsgi.I use this command to setup my server : mod_wsgi-express setup-server wsgi.py --port=80 --user test --group test -

问题 This is not a duplicate of Apache with virtualenv and mod_wsgi : ImportError : No module named 'django' since here I'm not using any virtualenv, and also I'm not trying to import another framework's module (such as django), but just a module in the same directory . Here is my setup: /var/www/test/app.py : import os, time, sys from bottle import route, run, template, default_app os.chdir(os.path.dirname(os.path.abspath(__file__))) import hello @route('/') def index(): return 'Hello world

问题 I know how to have a Python Bottle server: import os from bottle import route, template, default_app os.chdir(os.path.dirname(__file__)) @route('/hello') def hello(): return template('Hello world') application = default_app() run with WSGI, configured like this with Apache: <VirtualHost *:80> ServerName example.com <Directory /> AllowOverride All Require all granted </Directory> WSGIScriptAlias / /var/www/wsgi_test/app.wsgi </VirtualHost> Is it possible to do the WSGI configuration directly

问题 I am using this code to run django within twisted. from django.core.handlers.wsgi import WSGIHandler def wsgi_resource(): pool = threadpool.ThreadPool() pool.start() # Allow Ctrl-C to get you out cleanly: reactor.addSystemEventTrigger('after', 'shutdown', pool.stop) real_wsgi_app = WSGIHandler() def my_wsgi_wrapper(environ, start_response): environ['somekey'] = "somevalue" return real_wsgi_app(environ, start_response) wsgi_resource = wsgi.WSGIResource(reactor, pool, my_wsgi_wrapper) return