Using Flask to access Google's API

Using Flask to access Google's API

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Using Flask to access Google's API
Tag : python , By : usingtechnology
Date : December 05 2020, 12:22 PM

hope this fix your issue I'm trying to create a Flask app that talks to Google's API. However, I keep getting this error: , I think you are missing the leading slash from your google api route.

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How to access formData in flask sent using websockets?Flask-SocketIO

Tag : javascript , By : jazzyfox
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Instead of sending a FormData object, which is a client-side only construct, you should build a plain dictionary and send that, as all the data that is transferred back and forth in Socket.IO is serialized to JSON.
Then on the server, you will have a dict that is sent as an argument to your socket callback function. See this example for ideas on how to send form data to the server.

Possible to run flask web app on linux server without web access and access through tunnel?

Tag : python , By : Lior
Date : March 29 2020, 07:55 AM
help you fix your problem You can do this : Run your flask app or notebook on the remote server on any port. Say for example port 5000.
On your local machine run below command to establish the ssh tunnelling :
ssh -D 8123 -f -C -q -N username@remotesrrver
from flask import Flask
app = Flask(__name__)

def hello():
    return "Hello World!"

if __name__ == "__main__":
  app.run(host='', port=5901)

Access Flask instance when using Flask-Script

Tag : python , By : Ram
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , I am creating a flask instance using: , In __init__.py, initialize app and mongo to None.
app = None
mongo = None
import application

def create_app(config_name):
    app = Flask(__name__)
    return app

def create_mongo(app):
    return PyMongo(app)

app = create_app(Config.FLASK_CONFIG)

# bind to application package
application.app = app
application.mongo = create_mongo(app)
from application.setup import app

manager = Manager(app)
from application import mongo
cx, db = app.extensions['pymongo']['MONGO']

flask security separating access data on flask admin

Tag : python , By : Epora75
Date : March 29 2020, 07:55 AM
hop of those help? You need to add a permanent filter to your Room view by overriding get_query and get_count_query
For example (note I've named your Room class to RoomView as you already have a class called Room representing the database model):
class RoomView(UserAccess):

    def get_query(self):
        return self.session.query(self.model).filter(
            Room.user_id == current_user.id

    def get_count_query(self):
        return self.session.query(func.count('*')).select_from(self.model).filter(
            Room.user_id == current_user.id

    #  ... your code etc

Can't access Flask session variables as decorator arguments (in flask-SocketIO)

Tag : python , By : kbrust
Date : March 29 2020, 07:55 AM
I wish this help you Decorators execute when the module is imported, unlike the actual handler functions. At that early time there is no active request being handled, so there is no client. Since there is no actual client, session does not have any meaning.
I think you are not using namespaces in the expected way. You can't really have dynamic namespaces, you have to define a list of discrete namespaces and then implement handlers for all of them. For most applications this isn't very useful, so in most cases there is a single namespace for the entire application.
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