We Keep Coding

Python: Making a Flask Rest API Asynchronous and Deploying it

I have a python server that is currently keeping track of the location of all the buses in my university and generating predictions of arrivals to specific locations.
Now, I wanted to attach a lightweight REST API to this server but I have been running intro problems.
I tried using flask with the following code:
from flask import Flask, jsonify
from PredictionWrapper import *
import threading
class RequestHandler():
def __init__(self,predictionWrapper):
self.app = Flask(__name__)
self.predictor = predictionWrapper
self.app.debug = False
self.app.add_url_rule('/<route>/<int:busStop>','getSinglePrediction',self.getSinglePrediction)
t = threading.Thread(target=self.app.run, kwargs={'host':'0.0.0.0', 'port':80, 'threaded':True})
t.start()
def getSinglePrediction(self, route, busStop):
# TODO Get the actual prediction with given parameters
prediction = self.predictor.getPredictionForStop(route, busStop)
return jsonify({'busStop': busStop, 'prediction': prediction})
def getStopPrediction(self, busStop):
# TODO Get the actual prediction with given parameters
return jsonify({'busStop': busStop, 'prediction': 2})
def run(self):
self.app.run(host='0.0.0.0', port=80, threaded=True)
The problem is that I have been encountering the error below after about half a day of running the server. Note that no requests were made to the server around the time it failed with the following error:
ERROR:werkzeug: - - [01/May/2016 09:55:55] code 400, message Bad request syntax ('\x02\xfd\xb1\xc5!')
After investigating I believe I need to deploy to a WSGI production server. But I have no clue what it means in this specific approach given that 1)the flask server is being threaded in order to run the rest of the prediction application, and 2)I am using classes which none of the documentation uses.
Any help on how to setup the wsgi file with apache, gunicorn, or the technology of your choice would be appreciated. Also, any comments on a better approach on making a non-blocking REST API would be helpful.
Let me know if you need any further clarification!

Not sure if this can actually solve your problem but you can use the coroutine based web server gevent. They have a WSGI server that you can use if that's what you meant by saying that you need to deploy a WSGI production server.
If you want to implement the server to your flask application just do the following:
from gevent.pywsgi import WSGIServer
app = Flask(__name__)
http_server = WSGIServer(('', 5000), app)
http_server.serve_forever()
Gevent in general is a very powerful tool and by issuing context switches as necessary it can handle multiple clients very easily. Also, gevent fully supports flask.

First thing to do would be to put exception handling to deal with bad JSON request data (which maybe is what's happening) something like
def getSinglePrediction(self, route, busStop):
try:
prediction = self.predictor.getPredictionForStop(route, busStop)
return jsonify({'busStop': busStop, 'prediction': prediction})
except:
return jsonify({'busStop': 'error', 'prediction': 'error'})

Calling Python file in an Ajax Call

So I have established a pretty decent understanding of the simple architecture of an angularjs app, calling $http and posting to a php page, and receiving data back.
What I'm wondering, is how to do the same type of function with python. Is it possible to have python act the same, with self contained script files that accept post data and echo json back?
$username = $_POST['username'];
type variable assignment at the beginning of the script, and:
echo json_encode(response);
type response.
I'm wanting to use Python for some Internal Tools for my company, as it offers better libraries for remotely running powershell scripts (as the tools are all linux hosted) and overall just has libraries that fit my needs. I'm just having a difficult time finding a concise answer to how this could be set up.
---EDIT------
So I set up a quick example using the information below.
the angular:
var app = angular.module("api");
app.controller("MainController", ["$scope","$http",MainController]);
function MainController($scope,$http){
$http.post('/api',{test: "hello"})
.then(function(response){
console.log(response.data);
})
}
The flask:
from flask import Flask, request
import json
app = Flask(__name__)
#app.route('/api', methods=['POST', 'GET'])
def api():
if request.method == 'POST':
request.data
return 'You made it' # Just so I originally could see that the flask page
if __name__ == "__main__":
app.run()
I'm getting a 404 for that URL. If I change the angular to look at 'localhost:5000/api' (where my flask app is running),it gives me the error of "Unsupported URL Type".
I am seeing when I do the first case, it tries to look at http://localhost/api , which is correct! except for the port. Which is why I tried to specify the port.
Any suggestions for a next step?

Use flask.
You could host your app on a flask "server" and return the content you'd like too with a python processing.
http://flask.pocoo.org/
Use the documentation to setup a route where you'll POST your data using jquery or whatever, then on the route you can do your python stuff and return a JSON to your angular app if you need to.
from flask import request
#app.route('/test', methods=['POST', 'GET'])
def test():
if request.method == 'POST':
print request.data['your_field']
return your_json_data

Cloud9: Running a python server

In my Cloud9 IDE, which is running on Ubuntu I have encountered a problem in trying to reach my Python server externally. It's because their projects use a non-standard naming structure:
https://preview.c9users.io/{user}/{project}/
Changing the address to something like this, which is the default server address, doesn't help:
https://preview.c9users.io:8080/{user}/{project}/
I'm looking for a solution so I can run the following script or for a way to be able to combine HTML+JS+Python on Cloud9. The purpose of the server should be to respond to AJAX calls.
The Cloud9 server is Ubuntu-based, so there may be other ways to address this problem than just my script below.
import web
def make_text(string):
return string
urls = ('/', 'tutorial')
render = web.template.render('templates/')
app = web.application(urls, globals())
my_form = web.form.Form(
web.form.Textbox('', class_='textfield', id='textfield'),
)
class tutorial:
def GET(self):
form = my_form()
return render.tutorial(form, "Your text goes here.")
def POST(self):
form = my_form()
form.validates()
s = form.value['textfield']
return make_text(s)
if __name__ == '__main__':
app.run()

The server above actually runs and is available through URL in special format. It has been changed since earlier version, so I couldn't find it at first:
http://{workspacename}-{username}.c9users.io
Now I prefer to run it as a service (daemon) in the console window to execute additional scripts in the backend and test frontend functionality.

Flask Debugger not working under Windows

I am a newbie to Python attempting to experiment with sample code under Windows 8.1.
On http://flask.pocoo.org/docs/0.10/quickstart/ it says "if you enable debug support the server will reload itself on code changes, and it will also provide you with a helpful debugger".
I have added to the code app.run(debug=True) to the sample code on the above page. The server will now reload itself on code changes (as promised) but when I create a syntax error the "helpful debugger" does not show. Instead I get an error message in my command prompt.
Any ideas why? I suspect the answer might be here Can't enable debug mode in Flask but it is largely uninteligible to me.
So far I have tried restarting my machine and putting the code in different locations. I am not in a forked environment (as best as I know). For those that are curious my source code is shown below:
from flask import Flask
app = Flask(__name__)
from werkzeug.debug import DebuggedApplication
app.wsgi_app = DebuggedApplication(app.wsgi_app, True)
#app.route('/')
def hello_world():
return 'Hello World! #note the deliberate syntax error
if __name__ == '__main__':
app.debug = True
app.run()

The debugger is for debugging exceptions in a syntactically valid program. Try raising an exception in your code, visit the URL in a browser window, and you will see the very helpful debugger. See also http://flask.pocoo.org/snippets/21/
Syntax errors will be shown on the console as you have seen. That's how it works.

For Windows users
The page and resource that was selected as the answer in this post led to a broken page. However, try the code below.
Set debug mode using
$env:FLASK_ENV = "development"
Run app
flask run
This post solved it for me
Or
Set it directly from the app in your code.
if __name__ == '__main__':
app.debug = True
app.run(host='0.0.0.0', port=3000)

How to obtain values of parameters of get request in flask?

The answer that I found on the web is to use request.args.get. However, I cannot manage it to work. I have the following simple example:
from flask import Flask
app = Flask(__name__)
#app.route("/hello")
def hello():
print request.args['x']
return "Hello World!"
if __name__ == "__main__":
app.run()
I go to the 127.0.0.1:5000/hello?x=2 in my browser and as a result I get:
Internal Server Error
The server encountered an internal error and was unable to complete your request. Either the server is overloaded or there is an error in the application.
What am I doing wrong?

The simple answer is you have not imported the request global object from the flask package.
from flask import Flask, request
This is easy to determine yourself by running the development server in debug mode by doing
app.run(debug=True)
This will give you a stacktrace including:
print request.args['x']
NameError: global name 'request' is not defined

http://localhost:5000/api/iterators/opel/next?n=5
For something like the case before
from flask import Flask, request
n = request.args.get("n")
Can do the trick