I am learning to use the python bottle web framework. I have developed a simple app and I would like to suppress all the requests (urls) output on the terminal (though if there are any error msgs, their output should be ok). I tried setting the DEBUG to False, but that did not make any difference. Is there some other setting for this?
This works for some servers (including wsgiref):
bottle.run(..., quiet=True)
As mentionned by Helgi, the bottle dev server should'nt be used in production. With most of production server, you will not have these logs.
However, if you want to have a quiet dev server, something like this should work.
import bottle
class QuietWSGIRefServer(bottle.WSGIRefServer):
quiet = True
bottle.run(BOTTLE_APP, server=QuietWSGIRefServer)
I hope it helps
Related
I'm trying to create a simple REST Web Service using technology WSME reported here:
https://pypi.python.org/pypi/WSME
It's not clear, however, how to proceed. I have successfully installed the package WSME.0.6.4 but I don't understand how to proceed.
On the above link we can see some python code. If I wanted to test the code what should I do? I have to create a .py file? Where this file should be saved? Are there services to be started?
The documentation is not clear: it says "With this published at the / ws path of your application". What application? Do I need to install a Web Server?
Thanks.
You could use a full blown web server to run your application . For example Apache with mod_wsgi or uWSGI , but it is not always necessary .
Also you should choose a web framework to work with .
According WSME doc's it has support for Flask microframework out of the box , which is simple enough to start with .
To get started create a file with the following source code :
from wsgiref.simple_server import make_server
from wsme import WSRoot, expose
class MyService(WSRoot):
#expose(unicode, unicode)
def hello(self, who=u'World'):
return u"Hello {0} !".format(who)
ws = MyService(protocols=['restjson', 'restxml'])
application = ws.wsgiapp()
httpd = make_server('localhost', 8000, application)
httpd.serve_forever()
Run this file and point your web browser to http://127.0.0.1:8000/hello.xml?who=John
you should get <result>Hello John !</result> in response.
In this example we have used python's built in webserver which is a good choice when you need to test something out quickly .
For addition i suggest reading How python web frameworks and WSGI fit together
I am using Flask on Dreamhost. I (mostly) followed the instruction posted here https://mattcarrier.com/flask-dreamhost-setup/. I was able to successfully set up pages and direct urls to those pages, etc.
Now I want to be able to set a 404 page. I set the error handler using #app.errorhandler. The only problem is that it just returns a standard Apache 500 error page if I go to any page that doesn't have a URL set. Is this because of how Dreamhost has Apache setup? The public (static) folder is at the top level. How can I fix this?
You may want to try creating the file "missing.html" at the top level.
source: http://wiki.dreamhost.com/Creating_custom_error_pages
Edit:
To setup an error handling page from Flask itself, you do this:
from flask import render_template
#app.errorhandler(404)
def page_not_found(e):
return render_template('404.html'), 404
See: http://flask.pocoo.org/docs/patterns/errorpages/
My thought is that you are getting the 500 error because Flask/Python is crashing. Do you have debugging turned on? How is your routing setup?
I found the answer here - http://blog.tuxcoder.com/2011/9/8/dreamhost_python_wsgi/
If you plan to do any active development on the Dreamhost side, this step will simplify your efforts. Werkzeug comes with a really nice debugging engine, but it does not work with the Dreamhost configuration by default.
or example, if you have a simple Python coding error in your app, it will result in the following:
In this mode, you're out of luck. The only option is to startup a local server where you can go back and test the app. In some cases it might be a big effort just to replicate the bug.
With the Werkzeug Debugger enabled and patched, you'll get a much nicer output:
Here is what it says on the Dreamhost Wiki http://wiki.dreamhost.com/Flask (in passenger_wsgi.py:
# Uncomment next two lines to enable debugging
# from werkzeug.debug import DebuggedApplication
# application = DebuggedApplication(application, evalex=True)
I have a website (which running in Amazon EC2 Instance) running Python Bottle application with CherryPy as its front end web server.
Now I need to add another website with a different domain name already registered. To reduce the cost, I want to utilize the existing website host to do that.
Obviously, virtual host is the solution.
I know Apache mod_wsgi could play the trick. But I don't want to replace CherryPy.
I've googled a a lot, there are some articles showing how to make virtual hosts on CherryPy, but they all assume Cherrypy as Web Sever + Web application, Not CherrPy as Web server and Bottle as Application.
How to use CherrPy as Web server and Bottle as Application to support multiple virtual hosts?
As you mentioned, use VirtualHost. In the example cherrypy.Application instances are used, but any WSGI callable (e. g. Bottle app) will do.
perhaps you can simply put nginx as reverse proxy and configure it to send the traffic to the two domains to the right upstream (the cherryPy webserver).
Another idea would be to use Nginx (http://wiki.nginx.org/Main) with uWsgi(http://projects.unbit.it/uwsgi/) & (uWsgi-python) plug-in
uWsgi has a module named emperor that you can link vhosts(vassals) in, sort of.
i'm a newbie at this myself, so not necessarily an answer but rather a suggestion to check it out.
just a heads up, uWsgi and Nginx can be a hassle to get it to work, depending on your linux distro. Does work nicely with bottle, tested it myself.
hope it helps
jwalker's answer is pretty clear. In case any CherryPy newbie need whole script for reference, I post one below.
import cherrypy
from bottle import Bottle
import os
app1 = Bottle()
app2 = Bottle()
#app1.route('/')
def homePage():
return "========= home1 ==============="
#app2.route('/')
def homePage_2():
return "========= home2 ==============="
vhost = cherrypy._cpwsgi.VirtualHost(None,
domains={
'www.domain1.com': app1,
'www.domain2.com': app2,
}
)
cherrypy.tree.graft(vhost)
cherrypy.config.update({
'server.socket_host': '192.168.1.4',
'server.socket_port': 80,
})
cherrypy.engine.start()
cherrypy.engine.block()
you could make www.domain1.com and www.domain1.com point to one IP adress of you server, so it servers for 2 domain in one Web Server.
How to convert the web site develpoed in django, python into desktop application.
I am new to python and django can you please help me out
Thanks in Advance
I think you should just create an application that connects to the webserver. There is a good answer to getting RESTful API calls into your django application. This means you'd basically just be creating a new front-end for your server.
Using django-rest-interface
It doesn't make sense to rewrite the entire django application as a desktop application. I mean, where do you want to store the data?
For starters, you'll have to replace the web UI with a desktop technology like Tk/Tcl.
If you do that, you may not want to use HTTP as the protocol between the client and the services.
Django is a web framework. If you're switching to a desktop, you'll have to forego Django.
I would try to replicate the Django application functionality with the PyQt toolkit.
You can in fact embed web content in PyQt applications, with the help of QtWebKit. I would post some potentially useful links, but apparently I have too low a reputation to post more than one :)
There are two places you can go to try to decouple the view and put it into a new desktop app. First you can use the existing controller and model and adapt a new view to that. Second, you can use only the existing model and build a new view and controller.
If you haven't adhered closely enough to the MVC principles that you can detach the model from the rest of the application, you can simply rewrite the entire thing. if you are forced to go this route, bail on django and http entirely (as duffymo suggests above).
You have to also evaluate these solutions based upon performance requirements and "heaviness" of the services. If you have stringent performance requirements then relying on the HTTP layer just gets in the way, and providing a simple API into your model is the way to go.
There are clearly a lot of possibly solutions but this is the approach I would take to deciding what the appropriate one is...
It is possible to convert a django application to a desktop app with pywebview with some line of codes. Frist create a python file gui.py in directory where manage.py exists. install pywebview through pip, the write the code in gui.py
import os
import sys
import time
from threading import Thread
import webview
def start_webview():
window = webview.create_window('Hello world', 'http://localhost:8000/', confirm_close=True, width=900, height=600)
webview.start()
window.closed = os._exit(0)
def start_startdjango():
if sys.platform in ['win32', 'win64']:
os.system("python manage.py runserver {}:{}".format('127.0.0.1', '8000'))
# time.sleep(10)
else:
os.system("python3 manage.py runserver {}:{}".format('127.0.0.1', '8000'))
# time.sleep(10)
if __name__ == '__main__':
Thread(target=start_startdjango).start()
start_webview()
then run gui.py with the command python gui.py. IT will create a window like desktop app
There's a project called Camelot which seems to try to combine Django-like features on the desktop using PyQt. Haven't tried it though.
I am thinking about a similar Problem.
Would it be enough to habe a minimal PyQt Gui that enables you to present the djando-website from localhost (get rid of TCP/HTTPS on loop interface somehow) via QtWebkit?
All you seem to need is to have a minimal Python-Broser, that surfs the build in Webserver (and i guess you could even call Django directly for the html-payload without going over the HTTP/TCP layers).
I have django manage.py runserver in .bat file and a localhost bookmark bar in a browser and whola a django-desktop-app. Or make your own browser that opens localhost. Creating a web-browser with Python and PyQT
I'm on an app engine project where I'd like to put in a link to a Javascript test runner that I'd like to only exist when running the development server. I've made some experiments on a local shell with configuration loaded using the technique found in NoseGAE versus live on the 'App Engine Console' [1] and it looks to me like a distinction btw real instance and dev server is the presence of the module google.appengine.tools. Which lead me to this utility function:
def is_dev():
"""
Tells us if we're running under the development server or not.
:return:
``True`` if the code is running under the development server.
"""
try:
from google.appengine import tools
return True
except ImportError:
return False
The question (finally!) would be: is this a bad idea? And in that case, can anyone suggest a better approach?
[1] http://con.appspot.com/console/ (try it! very handy indeed)
The standard way to test for the development server is as follows:
DEBUG = os.environ['SERVER_SOFTWARE'].startswith("Dev")
Relying on the existence or nonexistence of a particular module - especially an undocumented one - is probably a bad idea.
I'd recommend doing it this way:
import os
def onDevServer():
return os.environ['SERVER_SOFTWARE'].find('Development') >= 0
This looks at the environment you're running in, and returns true if you're running on the development server. However, its a much cleaner way than checking an import, in my opinion.
I'm not a google app developer, but I wouldn't make this 100% dynamic, but also look at a value from a config file. I'm pretty sure you will be running into the problem, that you want to see this console on the prod system (google servers) or run your local version without the dev code (for testing).
To sum it up: Such a logic is fine for small stuff, like adding a debug link, but provide a way to overwrite it (e.g. by a configuration value)