Python sockets with flask and vue for continious serial data stream - python

So I have an flask application which is extended by the flask_socketio package.
I currently have a vue front end, that connects to an socket.
Now the output of the socket connection in the following:
I also see an connection in the terminal of the flask application:
127.0.0.1 - - [23/Aug/2019 21:07:11] "GET /socket.io/?EIO=3&transport=polling&t=Mo_u1wR HTTP/1.1" 200
However, I have on my code that when the socket connects, there should start an thread with while true, to continious receive the data.
This is the code I have:
from flask import Flask, jsonify
from flask_socketio import SocketIO, send, emit
import threading
from threading import Lock
import time
import controllers.gpsController
# configuration
DEBUG = True
# instantiate the app
app = Flask(__name__)
app.config.from_object(__name__)
thread = None
thread_lock = Lock()
async_mode = None
# app.config['SECRET_KEY'] = 'secret!'
# enable CORS
socketio = SocketIO(app, cors_allowed_origins='*', async_mode=async_mode)
def activate_gps():
ser = controllers.gpsController.open_serial_connection()
while True:
data = controllers.gpsController.readGPS(ser)
if data != None:
socketio.emit('fetch_gps_data', data, broadcast=True)
socketio.sleep(0.1)
# Use sockets here
#socketio.on('connect')
def start_get_data_thread():
global thread
with thread_lock:
if thread is None:
thread = socketio.start_background_task(target=activate_gps)
if __name__ == '__main__':
socketio.run(app, host='127.0.0.1', port=12345)
Anyone an idea what I'm doing wrong here? If I have te code of activate_gps() in another file and call that, I get the wanted output.

Related

Closing flask-socket io server programmatically

I am new to server development so please be kind...
I am developing a test application that starts a flask-socketio server and after interacting with a clients, it needs to shutdown and open another instance.
However this is not possible
I get error
File "C:\Python39\lib\site-packages\eventlet\convenience.py", line 78, in listen
sock.bind(addr)
OSError: [WinError 10048] Only one usage of each socket address (protocol/network address/port) is normally permitted
How can I programmatically shutdown the server?
I looked in answers here How to stop flask application without using ctrl-c and using a process indeed does the trick.
But I don't really want to have a separate process because sharing the variables between process is too tricky.
I also didn't understand from the same post how to send a request from the server to the server itself in order to shutdown the flask application.
This is an example of my code
import socketio
import eventlet
import eventlet.wsgi
from flask import Flask, render_template
import socket
import threading
import time
ip_addr=socket.gethostbyname(socket.gethostname())
appFlask = Flask(__name__)
sio = socketio.Server( ) #engineio_logger=True,logger=True)
# wrap Flask application with engineio's middleware
app = socketio.Middleware(sio, appFlask)
#sio.on('connect')
def connect(sid, environ):
print('connect ', sid)
#sio.on('message')
def message(sid, data):
print('message '+data, data)
#sio.on('disconnect')
def disconnect(sid):
print('disconnect ', sid)
#sio.on('result')
def result(sid,data):
print('result ', sid)
def worker1():
socket_port=3000
eventlet.wsgi.server(eventlet.listen((ip_addr, socket_port)), app)
if __name__ == '__main__':
sio.start_background_task(worker1)
# do some stuff and interact with the client
sio.sleep(2)
# how can I close the server so that I can do the following?
sio.start_background_task(worker1)
EDITED wit flask socket io functionality
import socketio
import eventlet
import eventlet.wsgi
from flask import Flask, render_template
import socket
import threading
import time
import requests
from flask import request
from flask_socketio import SocketIO
ip_addr=socket.gethostbyname(socket.gethostname())
socket_port=3000
app = Flask(__name__)
app.config['SECRET_KEY'] = 'secret!'
sio = SocketIO(app)
#app.route('/stop')
def stop():
sio.stop()
#sio.on('connect')
def connect(sid, environ):
print('connect ', sid)
#sio.on('message')
def message(sid, data):
print('message '+data, data)
#sio.on('disconnect')
def disconnect(sid):
print('disconnect ', sid)
#sio.on('result')
def result(sid,data):
print('result ', sid)
def worker1():
eventlet.wsgi.server(eventlet.listen((ip_addr, socket_port)), app)
if __name__ == '__main__':
eventlet_thr=sio.start_background_task(worker1)
# do some stuff and interact with the client
sio.sleep(2)
# now wait that the server is stopped
# invoke in a different process a request to stop
eventlet_thr.join()
# how can I close the server so that I can do the following?
sio.start_background_task(worker1)
You are using the eventlet web server is seems, so the question is how to stop the eventlet web server, Flask-SocketIO has nothing to do with the server.
As a convenience, Flask-SocketIO provides the stop() method, which you have to call from inside a handler. I'm not sure if that will work when the server is running on a thread that is not the main thread though, you'll have to test that out.
So basically what you need to do is add an endpoint that forces the server to exit, maybe something like this:
#app.route('/stop')
def stop():
sio.stop()
return ''
So then you can start and stop the server as follows:
if __name__ == '__main__':
thread = sio.start_background_task(worker1)
# do some stuff and interact with the client
requests.get('http://localhost:5000/stop')
thread.join()

How to exit the previously opened function created from socketio.start_background_task() when new connection is being made

Every time when I refresh the page from client side a new connection is made with the flask server and it runs the function 'backgroundFunction()' without exiting the recent opened function and the number increases as I refresh the page again and again.
from flask import Flask
from flask_socketio import SocketIO, send, emit
import socket
from time import sleep
import datetime
app = Flask(__name__)
app.config['SECRET_KEY'] = 'secret'
app.config['DEBUG'] = True
socketio = SocketIO(app , cors_allowed_origins="*" , async_mode = None , logger = False , engineio_logger = False)
def backgroundFunction():
while True:
data = "I am Data"
socketio.emit('data', data, broadcast=True)
socketio.sleep(2)
#socketio.on('connect')
def socketcon():
print('Client connected')
socketio.start_background_task(backgroundFunction)
if __name__ == ("__main__"):
socketio.run(app, port=5009)
Look at the example code in the Flask-SocketIO repository to learn one possible way to implement a background job that starts the first time an event is triggered.
Code is here. Here is the relevant excerpt:
thread = None
thread_lock = Lock()
def background_thread():
"""Example of how to send server generated events to clients."""
count = 0
while True:
socketio.sleep(10)
count += 1
socketio.emit('my_response',
{'data': 'Server generated event', 'count': count})
#socketio.event
def connect():
global thread
with thread_lock:
if thread is None:
thread = socketio.start_background_task(background_thread)

Flask socketIO connection established but not routed

For my project I have to connect one socketIO backend to another. For this I am using Flask-socketio and socketio-client. The code for both is the following:
CLIENT:
from socketIO_client import SocketIO, LoggingNamespace
ip = '192.168.1.41'
port = 8090
def handle_aaa_response():
print('response')
socketIO = SocketIO(ip, port)
socketIO.on('pingSuccess', on_aaa_response)
socketIO.wait(seconds=1)
SERVER:
from flask import Flask, render_template, jsonify, Response
from flask_socketio import SocketIO, emit
TRACE_LIBRARIES = False
HOST = '0.0.0.0'
WEB_PORT = 8090
USE_PIN = False
def handle_connect():
print('hello world')
emit('pingSuccess')
app = Flask(__name__)
app.config['SECRET_KEY'] = 'secret!'
app.config['DEBUG'] = True
socketio = SocketIO(app, cors_allowed_origins="*")
socketio.on('connect', handle_connect)
try:
socketio.run(app,
host=HOST,
port=WEB_PORT,
log_output=True)
except KeyboardInterrupt:
print('*** User raised KeyboardInterrupt')
exit()
When i run the client and server, the server only logs the following:
(4743) accepted ('192.168.1.33', 53500)
192.168.1.33 - - [21/Oct/2020 15:48:31] "GET /socket.io/?EIO=3&transport=polling&t=1603291711742-0 HTTP/1.1" 200 371 0.005033
(4743) accepted ('192.168.1.33', 53502)
This means the server is accepting the connection from the client, but not routing to the correct route on the server.
I want to know how I can change this so it gets to the correct route and prints "hello world:
Contrary to socketio.on() from the regular socketIO_client package you use in your client script, flask_socketio uses .on() as a decorator.
So to add a callback to an event in flask_socketio, you would need to change the following:
...
socketio = SocketIO(app, cors_allowed_origins="*")
#socketio.on('connect')
def handle_connect():
print('hello world')
emit('pingSuccess')
...
Server Side
#socketio.on('connect')
def test_connect():
print('hello world')
emit('pingSuccess')

Testing flask_sockets app

I am building an app using flask_sockets library. How do I test it?
Here is a code sample I want to write unit tests for:
import flask
from flask import Flask
from flask_sockets import Sockets
app = Flask(__name__)
sockets = Sockets(app)
ws_conns = []
#sockets.route('/echo')
def echo_socket(ws):
#on connect
ws_conns.append(ws)
#while connected
while True:
# message = ws.receive()
# if message is None:
# #socket has closed/errored
# break
for c in ws_conns:
c.send('hello and goodbye!')
#disconnected
ws_conns.remove(ws)
ws.close()
I have using this code from this git repo.

gevent-socketio send message from thread

I would like to use gevent-socketio to send messages from a worker thread and update all connected clients on the status of the job.
I tried this:
from flask import Flask, render_template
from flask.ext.socketio import SocketIO, send, emit
import threading
import time
app = Flask(__name__)
app.config['SECRET_KEY'] = 'secret!'
socketio = SocketIO(app)
#socketio.on('message')
def handle_message(message):
send(message, broadcast=True)
#app.route('/')
def index():
return render_template('index.html')
def ping_thread():
while True:
time.sleep(1)
print 'Pinging'
send('ping')
if __name__ == '__main__':
t = threading.Thread(target=ping_thread)
t.daemon = True
t.start()
socketio.run(app)
And it gives me this error:
RuntimeError: working outside of request context
How do I send messages from a function that doesn't have the #socketio.on() decorator? Can I use gevent directly to send messages to socketio?
From this section of the documentation:
Sometimes the server needs to be the originator of a message. This can be useful to send a notification to clients of an event that originated in the server. The socketio.send() and socketio.emit() methods can be used to broadcast to all connected clients:
def some_function():
socketio.emit('some event', {'data': 42})
This emit is not from from flask.ext.socketio import SocketIO, send, but instead called on your socketio variable from socketio = SocketIO(app). Had you done socketio_connection = SocketIO(app), then you'd be calling socketio_connection.emit() to broadcast your data.

Categories

Resources