Following some advice that I found here I am trying to use Flask as a web interface for an application that runs with twisted.
As suggested in Flask documentation I created a "templates" directory which is at the same level as my script but when I launch the server I get the following error:
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.
When I do not try to load a template and just write a string in the request it works fine. This is what makes me think it is related to the load of the template.
from twisted.internet import reactor
from twisted.web.resource import Resource
from twisted.web.wsgi import WSGIResource
from twisted.internet.threads import deferToThread
from twisted.web.server import Site, NOT_DONE_YET
from flask import Flask, request, session, redirect, url_for, abort, \
render_template, flash
app= Flask(__name__)
app.config.from_object(__name__)
#app.route('/login', methods= ['GET', 'POST'])
def login():
return render_template('login.html', error= error)
if __name__ == '__main__':
root = WSGIResource(reactor, reactor.getThreadPool(), app)
factory = Site(root)
reactor.listenTCP(8880, factory)
reactor.run()
Some frameworks will change directory from your current working directory when they are run in daemon mode, and this might very well be the case here.
Flask, since 0.7, has supported passing a template_folder keyword argument when calling Flask, so you could try:
import os
tmpl_dir = os.path.join(os.path.dirname(os.path.abspath(__file__)), 'templates')
The following is a shorter version that will work just fine:
tmpl_dir = os.path.join(os.path.dirname(__file__), 'templates)
# ...
app = Flask('myapp', template_folder=tmpl_dir)
You can feed Jinja2 with a default templates directory (as written here) like this :
import jinja2
app = Flask(__name__)
app.jinja_loader = jinja2.FileSystemLoader('path/to/templates/directory')
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
I want to set a basepath for my flask application. I have mentioned one example below.
basepath = 'http://localhost:3000/api'
i have two api call one is GET and other one is POST .
from flask import Flask
from flask_restful import Api
app = Flask(__name__)
api = Api(app)
api.add_resource(CreateUser, "/v1/user/create/")
api.add_resource(CreateUser, "/v1/user/details")
class CreateUser(Resource):
def post(self):
# Code for creating a user
def get(self):
# Code for get the details of user.
So here, if i want to create the user then my url will be http://localhost:3000/api/v1/user/create/
so same for GET also . So how do i achieve this ?
Initialize your Api with the path prefix:
from flask import Flask
from flask_restful import Api
app = Flask(__name__)
api = Api(app, "/api")
...
You can't change the host and port this way, you'll have to run flask with parameters:
flask run --host=127.0.0.1 --port=3000
Or you could do
from flask import Flask
from flask_restful import Resource, Api
app = Flask(__name__)
api = Api(app, "/api")
...
if __name__ == "__main__":
app.run(host="127.0.0.1", port="3000")
Please keep in mind this is not intended for production environments, only for local testing. Please see https://flask.palletsprojects.com/en/1.1.x/tutorial/deploy/ for using in a production environment.
If you want to get those values from basepath, one option would be purl:
url = purl.URL('http://localhost:3000/api')
url.host() # --> "localhost"
url.port() # --> 3000
url.path() # --> "/api"
I've used perfect Flask-SocketIO library with Python 3 for couple of months. Everything worked as expected until the last couple of days.
All works fine, if namespace for connection to websocket server stay default /socket.io. But I'm geting an error now if I trying to change namespace for connection to python flask-socketio backend.
My app.py:
from flask import Flask, session, request, render_template, copy_current_request_context
from flask_cors import CORS, cross_origin
import flask_socketio as socketio
import ujson
async_mode = 'threading'
namespace = '/mynamespace'
app = Flask(__name__)
CORS(app)
app.config['SECRET_KEY'] = 'secret!'
sio = socketio.SocketIO(app, async_mode=async_mode)
#sio.on('connect', namespace=namespace)
def connect():
logging.info('Connected')
#sio.on('disconnect', namespace=namespace)
def disconnect():
logging.info('Disconnected')
#app.route("/home/index")
def home():
return render_template('index.html',
async_mode=sio.async_mode)
I'm using ./main.py to run the server, main.py contains:
from app import app, sio
if __name__ == "__main__":
sio.run(app, debug=False)
My template/index.html contains ton of code js, but I think most valuable I loading the socketio from cdn in a head:
<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/socket.io/1.3.5/socket.io.min.js"></script>
... and I using connect with custom namespace path, as in docs:
namespace = '/mynamespace/socket.io';
var socket = io.connect(location.protocol + '//' + document.domain + ':' + location.port,
{path: namespace});
socket.on('connect', function() {
socket.emit('Connected to server');
})
As I understand, By default socketio library trying to connect to backend with emitting connect message to namespace. During loading 'index.html' template on '/home/index' route, logging the errors to console:
Flask server also gives and 404 error:
My best guess: at this moment it looks like something changed in client-side JS library or in chrome browser itself(few days ago I updated Chrome).
Maybe I just understood wrong one small detail. I really appreciate some help with this problem.
Stack versions:
Python 3.7.2,
Flask 1.0.2,
Flask-SocketIO 3.3.1,
socketio.min.js
1.3.5,
Google Chrome 77.0.3865.90 (64 bit)
You are confusing namespace with path, which are completely different things. The path is the endpoint URL where the Socket.IO server is listening. The namespace is a protocol feature of Socket.IO that allows multiplexing of multiple logical connections into a single physical connection.
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.
All
I have a flask_rest plus api over flask app in python. I want to deploy the same using the uwsgi format on nginx server. I searched for quite a long but didn't found the write solution.Any leads would be helpful..
Thanks in advance.
Below is the code snippet
import datetime
from fileinput import filename
import json
import os
from flask import request
from flask_restplus import Resource, Api
from werkzeug.utils import secure_filename
from APP.Application import app
from ApiConfig.ObjectConfig import Trainer, cData, Tmodal, Amodal, Cmodal
from ApiConfig.SA_utility import sentiment_sorting
from ApiConfig.appFileConfig import UPLOAD_FOLDER
from Classification.Classifier import classify
from DataBase.DB import db
from ExcelParser_DataFrame.Excel_Parser import ExcelDataFrame
from Vectorizer.Data_vector import vectorizer
from Mapper.UI_Map import UI_Map
from werkzeug.contrib.fixers import ProxyFix
api = Api(app)
app.wsgi_app = ProxyFix(app.wsgi_app)
#api.route('/sentiment-analysis/trainer')
class Upload_File(Resource):
def post(self):
..............
if __name__ == '__main__':
db.init_app(app)
db.create_all()
app.run(host='0.0.0.0')
My wsgi.py file looks like this
import sys
sys.path.insert(0, "/home/gulshan/Downloads/SA_PANDAS_SGDC/src")
from APP.Application import app
application = app
i tried using this wsgi file and started the server by firing the below command on my linux machine
uwsgi --http :5000 --wsgi-file wsgi.py --callable app --processes 4 --threads 2 --stats 127.0.0.1:9191
when i tried accessing the url using Postman then i get error as shown below:-
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<title>404 Not Found</title>
<h1>Not Found</h1>
<p>The requested URL was not found on the server. If you entered the URL manually please check your spelling and try again.</p>
My APP.Application file looks like this:-
from flask import Flask
from ApiConfig import Settings as settings
app = Flask(__name__)
#app.config['SERVER_NAME'] = settings.FLASK_SERVER_NAME
app.config['SQLALCHEMY_DATABASE_URI'] = settings.SQLALCHEMY_DATABASE_URI
app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = settings.SQLALCHEMY_TRACK_MODIFICATIONS
app.config['SWAGGER_UI_DOC_EXPANSION'] = settings.RESTPLUS_SWAGGER_UI_DOC_EXPANSION
app.config['RESTPLUS_VALIDATE'] = settings.RESTPLUS_VALIDATE
app.config['RESTPLUS_MASK_SWAGGER'] = settings.RESTPLUS_MASK_SWAGGER
app.config['ERROR_404_HELP'] = settings.RESTPLUS_ERROR_404_HELP
app.config['SQLALCHEMY_POOL_SIZE'] = settings.SQLALCHEMY_POOL_SIZE
#app.config['SQLALCHEMY_POOL_TIMEOUT'] = settings.SQLALCHEMY_POOL_TIMEOUT
app.config['SQLALCHEMY_POOL_RECYCLE'] = settings.SQLALCHEMY_POOL_RECYCLE
app.config['SQLALCHEMY_MAX_OVERFLOW'] = settings.SQLALCHEMY_MAX_OVERFLOW
Got a solution working for me although it seems to be weird way of doing this:-
Changed my uwsgi.py
import sys
sys.path.insert(0, "/home/gulshan/Downloads/SA_PANDAS_SGDC/src")
from APP.Application import app
from flask import Flask
from SA_APP.SA import appnew
application = app
application = appnew
and added this line in my first code snippet file containing main call
api = Api(app)
appnew = app.