Per suggestions on SO/SF and other sites, I am using CherryPy as the WSGI server to launch multiple instances of a Python web server I built with Flask. Each instance runs on its own port and sits behind Nginx. I should note that the below does work for me, but I'm troubled that I have gone about things the wrong way and it works "by accident".
Here is my current cherrypy.conf file:
[global]
server.socket_host = '0.0.0.0'
server.socket_port = 8891
request.dispatch: cherrypy.dispatch.MethodDispatcher()
tree.mount = {'/':my_flask_server.app}
Without diving too far into my Flask server, here's how it starts:
import flask
app = flask.Flask(__name__)
#app.route('/')
def hello_world():
return "hello"
And here is the command I issue on the command line to launch with Cherryd:
cherryd -c cherrypy.conf -i my_flask_server
Questions are:
Is wrapping Flask inside CherryPy still the preferred method of using Flask in production? https://stackoverflow.com/questions/4884541/cherrypy-vs-flask-werkzeug
Is this the proper way to use a .conf file to launch CherryPy and import the Flask app? I have scoured the CherryPy documentation, but I cannot find any use cases that match what I am trying to do here specifically.
Is the proper way to launch multiple CherryPy/Flask instances on a single machine to execute multiple cherryd commands (daemonizing with -d, etc) with unique .conf files for each port to be used (8891, 8892, etc)? Or is there a better "CherryPy" way to accomplish this?
Thanks for any help and insight.
I can't speak for Flask, but I can for CherryPy. That looks like the "proper way"...mostly. That line about a MethodDispatcher is a no-op since it only affects CherryPy Applications, and you don't appear to have mounted any (just a single Flask app instead).
Regarding point 3, you have it right. CherryPy allows you to run multiple Server objects in the same process in order to listen on multiple ports (or protocols), but it doesn't have any sugar for starting up multiple processes. As you say, multiple cherryd commands with varying config files is how to do it (unless you want to use a more integrated cluster/config management tool like eggmonster).
Terminology: Mounting vs Grafting
In principle this is a proper way to serve a flask app through cherrypy, just a quick note on your naming:
It is worth noting here that tree.mount is not a configuration key by itself - tree will lead to cherrypy._cpconfig._tree_config_handler(k, v) being called with the arguments 'mount', {'/': my_flask_server.app}.
The key parameter is not used at all by the _tree_config_handler so in your config "mount" is just an arbitrary label for that specific dict of path mappings. It also does not "mount" the application (it's not a CherryPy app after all). By that I mean, it does not cherrypy.tree.mount(…) it but rather cherrypy.tree.grafts an arbitrary WSGI handler onto your "script-name" (paths, but in CherryPy terminology) namespace.
Cherrypy's log message somewhat misleadingly says "Mounted <app as string> on /"]
This is a somewhat important point since with graft, unlike mount, you cannot specify further options such as static file service for your app or streaming responses on that path.
So I would recommend changing the tree.mount config key to something descriptive that does not invite reading too much semantics about what happens within CherryPy (since there is the cherrypy.tree.mount method) due to that config. E.g., tree.flask_app_name if you're just mapping that one app in that dict (there can be many tree directives, all of them just getting merged into the paths namespace) or tree.wsgi_delegates if you map many apps in that dict.
Using CherryPy to serve additional content without making an app of it
Another side note, if you want cherrypy to e.g. provide static file service for your app, you don't have to create a boilerplate cherrypy app to hold that configuration. You just have to mount None with the appropriate additional config. The following files would suffice to have CherryPy to serve static content from the subdirectory 'static' if they are put into the directory where you launch cherryd to serve static content (invoke cherryd as cherryd -c cherrypy.conf -i my_flask_server -i static:
static.py
import cherrypy
# next line could also have config as an inline dict, but
# file config is often easier to handle
cherrypy.tree.mount(None, '/static-path', 'static.conf')
static.conf
# static.conf
[/]
tools.staticdir.on = True
tools.staticdir.root = os.getcwd()
tools.staticdir.dir = 'static'
tools.staticdir.index = 'index.html'
Related
Gurus, Wizards, Geeks
I am tasked with providing Python Flask apps (more generally, webapps written in python) a way to reload properties on the fly.
Specifically, my team and I currently deploy python apps with a {env}.properties file that contains various environment specific configurations in a key value format (yaml for instance). Ideally, these properties are reloaded by the app when changed. Suppose a secondary application existed that updates the previously mentioned {env}.properties file, the application should ALSO be able to read and use new values.
Currently, we read the {env}.properties at startup and the values are accessed via functions stored in a context.py. I could write a function that could periodically update the variables. Before starting an endeavor like this, I thought I would consult the collective to see if someone else has solved this for Django or Flask projects (as it seems like a reasonable request for feature flags, etc).
One such pattern is the WSGI application factory pattern.
In short, you define a function that instantiates the application object. This pattern works with all WSGI-based frameworks.
The Flask docs explain application factories pretty well.
This allows you to define the application dynamically on-the-fly, without the need to redeploy or deploy many configurations of an application. You can change just about anything about the app this way, including configuration, routes, middlewares, and more.
A simple example of this would be something like:
def get_settings(env):
"""get the (current, updated) application settings"""
...
return settings
def create_app(env: str):
if env not in ('dev', 'staging', 'production'):
raise ValueError(f'{env} is not a valid environment')
app = Flask(__name__)
app.config.update(get_settings(env))
return app
Then, you could set FLASK_APP environment variable to something like "myapp:create_app('dev')" and that would do it. This is also the same way you could specify this for servers like gunicorn.
The get_settings function should be written to return the newest settings. It could even do something like retrieve settings from an external source like S3, a config service, or anything.
I'm working on an App Engine project (Python) where we'd like to make certain changes to the app's behavior when debugging/developing (most often locally). For example, when debugging, we'd like to disable our rate-limiting decorators, turn on the debug param in the WSGIApplication, maybe add some asserts.
As far as I can tell, App Engine doesn't naturally have any concept of a global dev-mode or debug-mode, so I'm wondering how best to implement such a mode. The options I've been able to come up with so far:
Use google.appengine.api.app_identity.get_default_version_hostname() to get the hostname and check if it begins with localhost. This seems... unreliable, and doesn't allow for using the debug mode in a deployed app instance.
Use os.environ.get('APPLICATION_ID') to get the application id, which according to this page is automatically prepended with dev~ by the development server. Worryingly, the very source of this information is in a box warning:
Do not get the App ID from the environment variable. The development
server simulates the production App Engine service. One way in which
it does this is to prepend a string (dev~) to the APPLICATION_ID
environment variable, which is similar to the string prepended in
production for applications using the High Replication Datastore. You
can modify this behavior with the --default_partition flag, choosing a
value of "" to match the master-slave option in production. Google
recommends always getting the application ID using get_application_id,
as described above.
Not sure if this is an acceptable use of the environment variable. Either way it's probably equally hacky, and suffers the same problem of only working with a locally running instance.
Use a custom app-id for development (locally and deployed), use the -A flag in dev_appserver.py, and use google.appengine.api.app_identity.get_application_id() in the code. I don't like this for a number of reasons (namely having to have two separate app engine projects).
Use a dev app engine version for development and detect with os.environ.get('CURRENT_VERSION_ID').split('.')[0] in code. When deployed this is easy, but I'm not sure how to make dev_appserver.py use a custom version without modifying app.yaml. I suppose I could sed app.yaml to a temp file in /tmp/ with the version replaced and the relative paths resolved (or just create a persistent dev-app.yaml), then pass that into dev_appserver.py. But that seems also kinda dirty and prone to error/sync issues.
Am I missing any other approaches? Any considerations I failed to acknowledge? Any other advice?
In regards to "detecting" localhost development we use the following in our applications settings / config file.
IS_DEV_APPSERVER = 'development' in os.environ.get('SERVER_SOFTWARE', '').lower()
That used in conjunction with the debug flag should do the trick for you.
In my development of Android and Java applications I have been using PHP scripts to interact with an online MySQL database, but now I want to migrate to Python.
How does one run Python scripts on a web server? In my experience with PHP, I have been saving my files under /var/www folder in a Linux environment. Then I just call the file later with a URL of the path. Where do I save my Python scripts?
You can use Flask to run webapps.
The simple Flask app below will help you get started.
from flask import Flask, jsonify
app = Flask(__name__)
#app.route('/sampleurl' methods = ['GET'])
def samplefunction():
#access your DB get your results here
data = {"data":"Processed Data"}
return jsonify(data)
if __name__ == '__main__':
port = 8000 #the custom port you want
app.run(host='0.0.0.0', port=port)
Now when you hit http://your.systems.ip:8000/sampleurl you will get a json response for you mobile app to use.
From within the function you can either do DB reads or file reads, etc.
You can also add parameters like this:
#app.route('/sampleurl' methods = ['GET'])
def samplefunction():
required_params = ['name', 'age']
missing_params = [key for key in required_params if key not in request.args.keys()]
if len(missing_params)==0:
data = {
"name": request.argv['name'],
"age": request.argv['age']
}
return jsonify(data)
else:
resp = {
"status":"failure",
"error" : "missing parameters",
"message" : "Provide %s in request" %(missing_params)
}
return jsonify(resp)
To run this save the flask app in a file e.g. myapp.py
Then from terminal run python myapp.py
It will start the server on port 8000 (or as specified by you.)
Flask's inbuilt server is not recommended for production level use. After you are happy with the app, you might want to look into Nginx + Gunicorn + Flask system.
For detailed instruction on flask you can look at this answer. It is about setting up a webserver on Raspberry pi, but it should work on any linux distro.
Hope that helps.
Use a web application framework like CherryPy, Django, Webapp2 or one of the many others. For a production setup, you will need to configure the web server to make them work.
Or write CGI programs with Python.
On Apache the simplest way would be to write the python as CGI here is an example:
First create an .htaccess for the web folder that is serving your python:
AddHandler cgi-script .py
Options +ExecCGI
Then write python that includes some some cgi libraries and outputs headers as well as the content:
#!/usr/bin/python
import cgi
import cgitb
cgitb.enable()
# HEADERS
print "Content-Type:text/html; charset=UTF-8"
print # blank line required at end of headers
# CONTENT
print "<html><body>"
print "Content"
print "</body></html>"
Make sure the file is owned by Apache chown apache. filename and has the execute bit set chmod +x filename.
There are many significant benefits to actually using a web framework (mentioned in other answers) over this method, but in a localhost web server environment set up for other purposes where you just want to run one or two python scripts, this works well.
Notice I didn't actually utilize the imported cgi library in this script, but hopefully that will direct you to the proper resources.
Most web development in python happens using a web framework. This is different than simply having scripts on the server, because the framework has a lot more functionality, such as handling URL routing, HTML templating, ORM, user session management, CSRF protection, and a lot of other features. This makes it easier to develop web sites, especially since it promotes component reuse, in a OOP fashion.
The most popular python web framework is Django. It's a fully-featured, tighly-coupled framework, with lots of documentation available. If you prefer something more modular and easier to customize, I personally recommend Flask. There's also lots of other choices available.
With that being said, if all you really want is to run a single, simple python script on a server, you can check this question for a simple example using apache+cgi.
I have a Flask application that will run under Apache in production. I have some static files, but they require authenticated access. So using X-Sendfile seemed reasonable to speed up the file delivery after authentication:
flaskapp = flask.Flask()
flaskapp.use_x_sendfile = True
Then where I'm generating the response:
return flask.send_file(filepath)
It seems to work fine under Apache. The problem is when I run the development server:
# Use SharedDataMiddleware to deliver JS, CSS, icons, etc.
flaskapp.wsgi_app = SharedDataMiddleware(flaskapp.wsgi_app, {'/static': '/path/to/static'})
flaskapp.run(host='0.0.0.0', debug=True)
When I run it this way with use_x_sendfile = True, the X-Sendfile header shows up in the actual response and an empty file (0 byte) is delivered to the client. Turning off X-Sendfile (and forcing the browser to discard the cached file) fixes the issue, so it seems the Werkzeug server Flask runs isn't processing the X-Sendfile header. Is there a way to enable the development server to process X-Sendfile, or am I forced to turn this off during development?
The flask documentation mentions that the server must support the X-Sendfile, and werkzeug's dev server doesn't (there's no mention anywhere in the documentation and a grep through werkzeug's source for sendfile returned no matches).
It wouldn't probably be too hard to add this behaviour to the server, but I'm not really sure if supporting such advanced features is a priority for a developement server.
Btw, as additional keyword arguments to run are passed through to werkzeug's run_simple, you don't need to create the middleware yourself, you can just use:
flaskapp.run(host='0.0.0.0', debug=True, static_files={'/static': '/path/to/static'})
Using the Google App Engine Python API is there a way to access the name of the currently running application--i.e., the app name specified in your app.yaml file with application: foobar?
import os
appname = os.environ['APPLICATION_ID']
EDIT: I just noticed this because I got a new upvote on it today (shame on you, upvoter!), but this is no longer correct.
from google.appengine.api.app_identity import get_application_id
appname = get_application_id()
should be used. The value in os.environ will include a "s~" prefix for applications using the HR datastore and, by default, "dev~" on the development server. (os.environ should also be avoided entirely on App Engine anyway, since when concurrency support is added with the Python 2.7 runtime, use of os.environ won't be threadsafe and will allow data to leak from one request to another, although obviously the application ID itself would be the same for multiple requests to the same application at the same time...)