I have a Python script written up and the output of this script is a list.
Right now I need to get it online and make it accessible to others. I looked at Django , but then I realized that it may be kind of hard to create the UI. Is there any simple way to create a UI in Django and map it to an existing Python script.
Right now I using nltk, numpy, sqlite3 and things like that. Or is there a simpler way by which I can proceed?
In your case, Django is redundant.
You can use something smaller, Flask or maybe Aiohttp.
For example, all you need in aiohttp:
basic hmtl template
handler for one url (here you will call your script)
aiohttp webserver
The main idea:
Your server catch some url (for example /),
start your script, receives result,
respond with your html template (also render script result in it).
You can try creating a flask App.
Just do a pip install Flask and try the code below
from flask import Flask
import flask
import json
from flask import Response
app = Flask(__name__)
#app.route('/test',methods=['GET'])
def test():
'''
GET: Receives the request in /test route and returns a response containing {"response": [1,2,3]}
'''
my_list = [1,2,3]
resp = Response(response=json.dumps({"response": my_list}), status=200, mimetype='application/json')
return resp
if __name__ == '__main__':
app.run(debug=True, host='0.0.0.0', port=8082)
Then to test your app from your browser accessing
localhost:8082/test
or also through some app like postman
I would suggest looking into something like React for creating the UI. This way your UI would only make calls to your Flask server.
Related
I'm trying to serve some simple service using flask and flask_restx (a forked project of flask-restplus, that would be eventually served on AWS.
When it is served, I want to generate swagger page for others to test it easily.
from flask import Flask
from flask_restx import Api
from my_service import service_namespace
app = Flask(__name__)
api = Api(app, version='1.0')
api.add_namespace(service_namespace)
if __name__ == '__main__':
app.run(debug=True)
When I test it locally (e.g. localhost:5000), it works just fine. Problem is, when it is hosted on AWS, because it has a specific domain (gets redirected?) (e.g. my-company.com/chris-service to a container), the document page is unable to find its required files like css and so:
What I've looked and tried
Python (Flask + Swagger) Flasgger throwing 404 error
flask python creating swagger document error
404 error in Flask
Also tried adding Blueprint (albeit without knowing exactly what it does):
app = Flask(__name__)
blueprint = Blueprint("api", __name__,
root_path="/chris-service",
# url_prefix="/chris-service", # doesn't work
)
api = Api(blueprint)
app.register_blueprint(blueprint)
...
And still no luck.
Update
So here's more information as per the comments (pseudo, but technically identical)
Access point for the swagger is my-company.com/chris (with or without http:// or https:// doesn't make difference)
When connecting to the above address, the request URL for the assets are my-company.com/swaggerui/swagger-ui.css
You can access the asset in my-company.com/chris/swaggerui/swagger-ui.css
So I my resolution (which didn't work) was to somehow change the root_path (not even sure if it's the correct wording), as shown in What I've looked and tried.
I've spent about a week to solve this but can't find a way.
Any help will be greatful :) Thanks
Swagger parameters defined at apidoc.py file. Default apidoc object also created in this file. So if you want to customize it you have change it before app and api initialization.
In your case url_prefix should be changed (I recommend to use environment variables to be able set url_prefix flexibly):
$ export URL_PREFIX='/chris'
from os import environ
from flask import Flask
from flask_restx import Api, apidoc
if (url_prefix := environ.get('URL_PREFIX', None)) is not None:
apidoc.apidoc.url_prefix = url_prefix
app = Flask(__name__)
api = Api(app)
...
if __name__ == '__main__':
app.run()
Always very frustrating when stuff is working locally but not when deployed to AWS. Reading this github issue, these 404 errors on swagger assets are probably caused by:
Missing javascript swagger packages
Probably not the case, since flask-restx does this for you. And running it locally should also not work in this case.
Missing gunicorn settings
Make sure that you are also setting gunicorn up correctly as well with
--forwarded-allow-ips if deploying with it (you should be). If you are in a kubernetes cluster you can set this to *
https://docs.gunicorn.org/en/stable/settings.html#forwarded-allow-ips
According to this post, you also have to explicitly set
settings.FLASK_SERVER_NAME to something like http://ec2-10-221-200-56.us-west-2.compute.amazonaws.com:5000
If that does not work, try to deploy a flask-restx example, that should definetely work. This rules out any errors on your end.
I am creating a REST API using python flask. The API is ready and works on port number 8000 of my localhost. Now I intend to give this REST API a user friendly interface for which I decided to go with python - restplus. I thought of calling this service (running on 8000) internally from swagger application running on 5000
I was able to create the basic structure of the API (Swagger). The code for which looks like this:
import flask
from flask import Flask, request
from flask_restplus import Resource, Api
app = Flask(__name__)
api = Api(app)
#api.route('/HybridComparator/<string:url2>/<string:url1>')
class HybridComparator(Resource):
def get(self, url1, url2):
print(url1)
print(url2)
return url1 + ' ' + url2
if __name__ == '__main__':
app.run(debug=True)
The application as a whole runs seamlessly (with random strings as parameters) on port 5000. But when the URLs I pass are actual links, the application returns a response of 404 - Not found. Further to my investigation I realized the culprit being '/' embedded within the links I try to provide. Is there a way to handle URLs in particular?
Should I encode them before sending a request. (This will make my parameters look ugly). Is there something I am missing?
This is an entirely old question and I am sure you solved your problem by now.
But for new searchers, this may come in handy;
replace <string:url2>/<string:url1> with <path:url2>/<path:url1>
it seems that :
#api.route('/HybridComparator/<path:url2>/<path:url1>')
should fix it ,it fixes the 404 but i am getting only "http:/" part of the param
i'm learning to use python flask to complement angularJS. This is what i have on flask:
#!flask/bin/python
from flask import Flask, jsonify
app = Flask(__name__)
#app.route('/hola', methods=['GET'])
def get_tasks():
x="pepepepep"
return jsonify({'msj': x})
if __name__ == '__main__':
app.run(debug=True)
when i open in my browser:
http://127.0.0.1:5000/hola
i do indeed see the json object with that random message.
Now, when i go to angularJS, this is what i have on the controller:
function ControladorPrueba($scope,$http) {
$scope.mensaje="i";
$scope.ObtenerMsj= function(){
$http.get('http://localhost:5000/hola')
.success(function(data){
$scope.mensaje=data.msj;
})
.error(function(data,status){
$scope.mensaje = status;
});
};
}
The problem is that when this function is executed, the get always go on .error(...), any ideas of why this happens even when the service in flask works well when opened on the browser? am i missing something?
Any help appreciated. Thanks!
I suspect that this may be due to the Same-origin policy. If you're running your Flask app on localhost, and if you're serving your Angularjs files separately (for example, using the file:/// protocol), then the browser will actually forbid your Angular code from talking to the webservice for security reasons.
If this is the case, you have two solutions. You can either start serving your HTML/CSS/Javascript files from your Flask app, or by using something like Flask-Cors to explicitly set your server to allow external requests.
My web app assigns a subdomain to users and optionally allows them to use a custom domain. This works except when the user visits their custom domain for a route without including a trailing slash.
GET requests to this url works as expected: http://user.example.com:5000/book/12345/
GET requests to this url works as expected: http://custom.com:5000/book/12345/
GET requests to this url attempt to redirect, but fail: http://custom.com:5000/book/12345
Flask ends up redirecting the browser to this url which, of course, doesn't work: http://<invalid>.example.com:5000/book/12345/
Is there a different way that I should handle custom domains? Here's a complete minimal example to reproduce this. I have set custom.com, example.com. and user.example.com to point to 127.0.0.1 in my /etc/hosts file in my development environment so that Flask receives the request.
from flask import Flask
app = Flask(__name__)
server = app.config['SERVER_NAME'] = 'example.com:5000'
#app.route('/', subdomain="<subdomain>")
#app.route('/')
def index(subdomain=None):
return ("index")
#app.route('/book/<book_id>/', subdomain="<subdomain>")
#app.route('/book/<book_id>/')
def posts(post_id, subdomain=None):
return (book_id)
if __name__ == '__main__':
app.run(host='example.com', debug=True)
I'm not sure that's possible. host matching and subdomain matching are mutually exclusive (look at host matching parameter).
I'd love to be wrong though.
One way around this issue that I can think of is to use something in front of Flask (say nginx) that points custom.com to custom.com._custom.example.com or something like that. In your code you could create a custom url_for function that would recognize this as a custom domain. I would ask on the Flask mailing list as they would be able to give you a solid answer.
Is there a way to call a python function when a certain link is clicked within a html page?
Thanks
You'll need to use a web framework to route the requests to Python, as you can't do that with just HTML. Flask is one simple framework:
server.py:
from flask import Flask, render_template
app = Flask(__name__)
#app.route('/')
def index():
return render_template('template.html')
#app.route('/my-link/')
def my_link():
print 'I got clicked!'
return 'Click.'
if __name__ == '__main__':
app.run(debug=True)
templates/template.html:
<!doctype html>
<title>Test</title>
<meta charset=utf-8>
Click me
Run it with python server.py and then navigate to http://localhost:5000/. The development server isn't secure, so for deploying your application, look at http://flask.pocoo.org/docs/0.10/quickstart/#deploying-to-a-web-server
Yes, but not directly; you can set the onclick handler to invoke a JavaScript function that will construct an XMLHttpRequest object and send a request to a page on your server. That page on your server can, in turn, be implemented using Python and do whatever it would need to do.
Yes. If the link points to your web server, then you can set up your web server to run any kind of code when that link is clicked, and return the result of that code to the user's browser. There are many ways to write a web server like this. For example, see Django. You might also want to use AJAX.
If you want to run code in the user's browser, use Javascript.
There are several ways to do this, but the one that has worked best for me is to use CherryPy. CherryPy is a minimalist python web framework that allows you to run a small server on any computer. There is a very similiar question to yours on stackoverflow - Using the browser for desktop UI.
The code below will do what you want. Its example 2 from the CherryPy tutorial.
import cherrypy
class HelloWorld:
def index(self):
# Let's link to another method here.
return 'We have an important message for you!'
index.exposed = True
def showMessage(self):
# Here's the important message!
return "Hello world!"
showMessage.exposed = True
import os.path
tutconf = os.path.join(os.path.dirname(__file__), 'tutorial.conf')
if __name__ == '__main__':
# CherryPy always starts with app.root when trying to map request URIs
# to objects, so we need to mount a request handler root. A request
# to '/' will be mapped to HelloWorld().index().
cherrypy.quickstart(HelloWorld(), config=tutconf)
else:
# This branch is for the test suite; you can ignore it.
cherrypy.tree.mount(HelloWorld(), config=tutconf)
I personally use CherryPy in combination with several other modules and tools:
Mako (template library)
py2exe (convert into Windows executable)
GccWinBinaries (used in combination with py2exe)
I wrote an article about Browser as Desktop UI with CherryPy that introduces modules and tools used plus some further links that might help.
In addition to running Python scripts on a server, you can run Python scripts on the client-side using Skulpt.