I'm trying to use flask-socketio. I ran a simple example from the docs and it worked fine. I'm using a virtual environment
app.py
from flask import Flask, render_template
from flask_socketio import SocketIO
app = Flask(__name__)
app.config['SECRET_KEY'] = 'secret!'
socketio = SocketIO(app)
#socketio.on('message')
def handle_message(message):
print('received message: ' + message)
if __name__ == '__main__':
socketio.run(app, port=8000)
and in demo.py I use the socketio client library
demo.py
from socketIO_client import SocketIO, LoggingNamespace
def on_message_response(*args):
print('on_message_response', args)
with SocketIO('localhost', 8000, LoggingNamespace) as socketIO:
socketIO.emit('message', 'xxx', on_message_response)
socketIO.wait_for_callbacks(seconds=1)
I've installed eventlet and this basic example worked fine. When I ran I got the output received message: xxx However when I updated the app.py file and demo.py to this
app.py
from flask import Flask, render_template
from flask_socketio import SocketIO
app = Flask(__name__)
app.config['SECRET_KEY'] = 'secret!'
socketio = SocketIO(app)
#socketio.on('message')
def handle_message(name, age):
print('We Welcome {} who is {} years old'.format(name,age))
if __name__ == '__main__':
socketio.run(app, port=8000)
demo.py
from socketIO_client import SocketIO, LoggingNamespace
def on_message_response(*args):
print('on_message_response', args)
with SocketIO('localhost', 8000, LoggingNamespace) as socketIO:
socketIO.emit('message', {'name':'Dipanshu', 'age':25}, on_message_response)
socketIO.wait_for_callbacks(seconds=1)
It doesnt seem like the app.py reloads or is updating because I keep getting this
File "app.py", line 10, in handle_message
print('received message: ' + message) TypeError: must be str, not dict
I have restarted the app.py a few times but I get the same error. I also tried uninstalling eventlet and ran it using the Flask server like other Flask app but it was still the same.
This could be an issue with compiled Python files. I'm going to give you a couple of ideas to try:
My first suggestion is for you to delete all *.pyc files and __pycache__ directories that you have in your main application directory or in any sub-directories. This is going to remove all compiled Python files, and will force Python to recompile everything.
Alternatively, you can also take your two source files and move them to a brand new directory, and then run your server and client from the new directory. This should also clear any issues due to caching of compiled Python files.
Hope this helps!
Related
POST UPDATE
I tested this same code on a different machine and it worked perfectly so it is probably a problem with some dependencies or something similar.
from flask import Flask, render_template
from flask_socketio import SocketIO
app = Flask(__name__)
socketio = SocketIO(app)
#app.route('/')
def index():
return render_template("index.html")
#socketio.on('test')
def test(message):
print(message)
if __name__ == '__main__':
socketio.run(app,host='0.0.0.0',port=8080,debug=True)
I saw in the documentation that you can run the web and socket server just by using
the SocketIO.run method.
When I execute my app.py it runs the server but I cannot access with my browser to webpage.
I saw somewhere that using async_mode='eventlet' on the SocketIO constructor could help but it threw an error.
ValueError: Invalid async_mode specified
The following scenario:
I have a Raspberry Pi running as a server. Currently I am using a Python script with Flask and I can also access the Raspberry Pi from my PC. (The flask server runs an react app.)
But the function should be extended. It should look like the following:
2nd Python script is running all the time. This Python script fetches data from an external API every second and processes it. If certain conditions are met, the data should be processed and then the data should be communicated to the Python Flask server. And the Flask server then forwards the data to the website running on the computer.
How or which method is best to program this "interprocess communication". Are there any libraries? I tried Celery, but then it throws up my second Python script whenever I want to access the external API, so I don't know if this is the right choice.
What else would be the best approach? Threading? Direct interprocess communication?
If important, this is how my server application looks so far:
from gevent import monkey
from flask import Flask, render_template
from flask_socketio import SocketIO
monkey.patch_all()
app = Flask(__name__, template_folder='./build', static_folder='./build/static')
socket_io = SocketIO(app)
#app.route('/')
def main():
return render_template('index.html')
#socket_io.on('fromFrontend')
def handleInput(input):
print('Input from Frontend: ' + input)
send_time()
#socket_io.on('time')
def send_time():
socket_io.emit('time', {'returnTime': "some time"})
if __name__ == '__main__':
socket_io.run(app, host='0.0.0.0', port=5000, debug=True)
Well i found a solution for my specific problem i implemented it with a thread as follows:
import gevent.monkey
gevent.monkey.patch_all()
from flask import Flask, render_template
from flask_socketio import SocketIO
import time
import requests
from threading import Thread
app = Flask(__name__, template_folder='./build', static_folder='./build/static')
socket_io = SocketIO(app)
#app.route('/')
def main():
thread = Thread(target=backgroundTask)
thread.daemon = True
thread.start()
return render_template('index.html')
#socket_io.on('fromFrontend')
def handleInput(input):
print('Input from Frontend: ' + input)
#socket_io.on('time')
def send_time():
socket_io.emit('time', {'returnTime': 'hi frontend'})
def backgroundTask():
# do something here
# access socket to push some data
socket_io.emit('time', {'returnTime': "some time"})
if __name__ == '__main__':
socket_io.run(app, host='0.0.0.0', port=5000, debug=True)
I need help in debugging -the Same Origin Policy disallows reading the remote resource at https://some-domain.com. (Reason: CORS request did not succeed) in python flask-socketio error.
I am working on a chat application using python flask-socketio. In previously I have created that application in local and it works fine as expected, while I move the below code to the server it shows the above error. The client code runs in the https servers and server code also runs on the https server I don't know why that error shows.
I have attached my code below and please give a better solution to me.
server.py
import json
import os
from flask import Flask, render_template, request,session
from flask_socketio import SocketIO, send, emit
from datetime import timedelta,datetime
from flask_cors import CORS
app = Flask(__name__)
app.config['SECRET_KEY'] = 'secretkey'
app.config['DEBUG'] = True
app.config['CORS_HEADERS'] = 'Content-Type'
cors = CORS(app, resources={r"/*": {"origins": "*"}})
socketio = SocketIO(app)
users = {}
#app.before_request
def make_session_permanent():
session.permanent = True
app.permanent_session_lifetime = timedelta(minutes=1)
#app.route('/')
##cross_origin(origin='*',headers=['Content- Type','Authorization'])
def index():
return render_template('index.html')
#socketio.on('connect')
def connect():
print("connected");
#app.route('/orginate')
def orginate():
socketio.emit('server orginated', 'Something happened on the server!')
return '<h1>Sent!</h1>'
#socketio.on('username')
def receive_username(username):
users[username] = request.sid
#users.append({username : request.sid})
#print(users)
emit('userList', users, broadcast=True)
print('Username added!')
print(users)
if _name_ == '__main__':
socketio.run(app,host='xxx.xxx.xx.x',port=5001)
client.js
var socket = io.connect("https://xxx.xxx.xx.x:5001/",{secure:false});
Screenshot 1:
This screenshot explains the access-control-allow-orgin works fine for images under static folder in flask framework
Screenshot 2:
This screenshot explains there is no access-control-orgin for socket call
You are using Flask-CORS to set up CORS on your Flask routes. You are missing a similar set up for Flask-SocketIO:
socketio = SocketIO(app, cors_allowed_origins=your_origins_here)
You can use '*' as the value to allow all origins (which I do not recommend), or set a single origin as a string, or a list of origins as a list of strings.
I am trying to host a flask server from my windows computer so I can access it from external devices
I am using Flask/Python and have already tried a few things but can't get it to work
Tried running it on 0.0.0.0, port 33, 5000, etc. but I still can't access it this way
from flask import Flask, request, abort
app = Flask(__name__)
if __name__ == '__main__':
app.run(host='0.0.0.0', port=33)
When I then run the file I get:
Running on http://0.0.0.0:33/ (Press CTRL+C to quit)
But it isn't even running there, nor on any other way I can access it
I expect to be able to access my flask application and send requests to it by using my public IP address
What can I do here to make it work?
You have missed an important line in your code:
After the line
app = Flask(__name__)
You have to write the line:
#app.route('/')
We use the route() decorator to tell Flask what URL should trigger our function.
And then define a function that will tell what task to be performed in the web app hosted in the respective address.
The function might look something like this:
def hello_world():
return 'Hello, World!'
The complete code then will look like:
from flask import Flask
app = Flask(__name__)
#app.route('/')
def hello_world():
return 'Hello, World!'
if __name__ == '__main__':
app.run(host='0.0.0.0', port=33)
Hope this helps.
I'm trying to create a python Flask REST web API. Since Flask development server is not suitable for production, I tried to use cherrypy application server.
Following is the Flask app I tried to expose via cherrypy
from flask import Flask,request
from flask_restful import Api,Resource, reqparse
app= Flask(__name__)
api = Api(app)
class Main (Resource):
def get(self):
return "Hello Flask"
if __name__ == '__main__':
api.add_resource(Main, "/testapp/")
app.run(debug=True)
Following is the cherrypy script I have created
try:
from cheroot.wsgi import Server as WSGIServer, PathInfoDispatcher
except ImportError:
from cherrypy.wsgiserver import CherryPyWSGIServer as WSGIServer, WSGIPathInfoDispatcher as PathInfoDispatcher
from stack import app
d = PathInfoDispatcher({'/': app})
server = WSGIServer(('127.0.0.1', 8080), d)
if __name__ == '__main__':
try:
server.start()
print("started")
except KeyboardInterrupt:
server.stop()
I have saved this script as "run.py" in my project directory. When I run this it doesn't show any error, which made me to thin this is correct.
But unfortunately I cant access this using the url
Theoretically, url for this API should be some thing like follow
http://127.0.0.1:8080/testapp/
But it throws 404 with the message
"The requested URL was not found on the server. If you entered the URL manually please check your spelling and try again."
What am I doing wrong ?
The
api.add_resource(Main, "/testapp/")
in your file stack.py is not executed if the file is included from your run.py
as the condition
if __name__ == '__main__':
...
is not true (in the context of stack.py).
Moving the call to api.add_resource(...) to a position outside the if-main-condition (so it is always executed) should solve the issue.