I'm running this very simple application.py file, in debug mode, on my Mac.
import os
from flask import Flask, render_template
from flask_socketio import SocketIO, emit
app = Flask(__name__)
app.config["SECRET_KEY"] = "secret"
socketio = SocketIO(app)
#app.route("/")
def index():
return "hello"
if __name__ == '__main__':
socketio.run(app, debug=True)
It shows up on 127.0.0.1:5000. When I change return "hello" to return "goodbye" and refresh the page, nothing happens. When I try to return render_template(goodbye.html) from my templates directory, nothing happens. I even changed the route from '/' to '/bbbb' and nothing changed. I see a bunch of GET requests in my terminal, each with a status code of 200.
I've never had this problem with Flask, that is until I tried to use sockets.io. Any thoughts on what is happening?
Related
I am trying to deploy my flask application to aws app runner, locally everything works perfectly. But I can't figure out how to redirect to another page of my website in app runner
My code looks similar to this
from flask import Flask, url_for
from waitress import serve
app = Flask(__name__)
#app.route("/hello")
def hello():
return "Hello"
#app.route("/redirect_to")
def redirect_to():
return "Redirected successfully!"
#app.route("/redirect_from")
def redirect_from():
return redirect(url_for("redirect_to"))
if __name__ == "__main__":
serve(app, host="0.0.0.0", port=8000)
App runner provided "Default domain" that redirects all traffic to my app, that is running on 0.0.0.0:8000. When I request default-domain.awsapprunner.com/hello, it successfully redirects to 0.0.0.0:8000/hello, but when I try to request default-domain.awsapprunner.com/redirect_from page loads forever. I think it happens because my app redirects to 0.0.0.0, and app runner expects that all traffic comes to default-domain.awsapprunner.com but I am not sure
What is the best way to fix this problem?
from flask import Flask, url_for, redirect
from waitress import serve
app = Flask(__name__)
#app.route("/hello")
def hello():
return "Hello"
#app.route("/redirect_to")
def redirect_to():
return "Redirected successfully!"
#app.route("/redirect_from")
def redirect_from():
return redirect("http://YOUR_APP_URL.com/redirect_to")
if __name__ == "__main__":
serve(app, host="0.0.0.0", port=8000)
Usecase: I have a python flask app that runs background_function() before serving any requests on routes.
When I execute the flask app, I receive the error - RuntimeError: Working outside of application context. I receive the error since I try to get the application context before any request is served.
What is the best pythonic way to execute the background_function() in this example?
from flask import Flask
from download import Download
app = Flask(__name__)
app.config.from_pyfile('config.py')
# run backgroung function
Download.background_function()
#app.route('/')
def index():
return 'Welcome!'
if __name__ == '__main__':
app.run()
The config file
FILE_LOCATION = os.environ['FILE_LOCATION'] # "file/path/on/server"
# Many other variables are present in this file
The download file
from flask import current_app as app
class Download:
#staticmethod
def background_function():
file_path = app.config["FILE_LOCATION"]
# code to download file from server to local
return
Try this:
from flask import Flask
from download import Download
app = Flask(__name__)
#app.route('/')
def index():
return 'Welcome!'
if __name__ == '__main__':
Download.background_function()
app.run()
the download file
from flask import current_app as app
class Download:
#staticmethod
def background_function():
print("testing")
given output:
testing
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
As you can see, the function runs first and prints testing and then runs the application.
I am trying to build some restful API's. When I try to segregate code into packages the service doesn't work and I get URL not found on the server. For examples:
Scenario 1 [Works fine as I have everything in main.py]
from flask import Flask, jsonify, request
app = Flask(__name__)
#app.route('/echo', methods=['POST'])
def echo():
message = request.get_json().get('message', '')
return jsonify({'message': message})
if __name__ == '__main__':
app.run(host='127.0.0.1', port=8080, debug=True)
Now when I try to segregate the code into different packages, it just doesn't work. For example:
Scenario 2 [Doesn't work as the code is in different packages]
I am initializing the app in api/restful.py
from flask import Flask, jsonify, request
app = Flask(__name__)
Then created a service in api/endpoints/service.py
from api.restplus import app, jsonify, request
#app.route('/echo', methods=['POST'])
def echo():
message = request.get_json().get('message', '')
return jsonify({'message': message})
Finally in main.py
from api.restplus import app
if __name__ == '__main__':
app.run(host='127.0.0.1', port=8080, debug=True)
It seems like the service is not visible to the app when I put it in a different package. Please advise.
Assuming that the issue you get is that flask does not see your service it looks like nothing is importing your service code once you split your code.
Simply modify your main.py file to look like this to fix it:
from api.restplus import app
import api.endpoints.service
if __name__ == '__main__':
app.run(host='127.0.0.1', port=8080, debug=True)
Hope this helps !
You may want to do this way Or I would suggest, if there are less routes try to have everthing in one file.
from yourfile import app
if __name__ == '__main__':
app.run(host='127.0.0.1', port=8080, debug=True)
In yourfile.py
from flask import Flask, jsonify, request
app = Flask(__name__)
#app.route('/echo', methods=['POST'])
def echo():
message = request.get_json().get('message', '')
return jsonify({'message': message})
I made the following file yesterday.
# import flask
from flask import Flask
from flask import render_template
from flask import request
app = Flask(__name__)
# create url & function mapping for root or /
#app.route('/')
def index():
return "Hello from Flask"
# create another mapping name /hello
#app.route('/hello')
def hello():
myName = "kayak"
return "Hello again !!" + myName
# create mapping for /myprofile
#app.route('/myprofile')
def showmyprofile():
return render_template('myprofile.html')
# create mapping for /myprofile
#app.route('/addprofileform')
def addprofileform():
return render_template('myprofileform.html')
# create a mapping for /addprofile
#app.route('/addprofile')
def addprofile():
myname = request.args.get('myname')
state_of_residence = request.args.get('state_of_residence')
return render_template('myprofile.html', html_page_name=myname,
html_page_state_of_residence=state_of_residence)
if __name__== '__main__':
app.run()
Then I made the following file today.
from flask import Flask
app = Flask(__name__)
#app.route('/')
def index():
return 'This is the homepage'
if __name__ == "__main__":
app.run(debug=True)
I thought
app.run(debug=True)
would work to clear the old data, but I doesn't and http://127.0.0.1:5000/ page keeps showing "Hello from Flask".
How do I fix this?
Just clear the cache in your browser and try running it again.
Here's how to clear your cache in some browsers:
Firefix->https://support.mozilla.org/en-US/kb/how-clear-firefox-cache
Chrome->https://support.google.com/accounts/answer/32050?co=GENIE.Platform%3DDesktop&hl=en
You can export the FLASK_ENV environment variable and set it to development before running the server
export FLASK_ENV=development
flask run
This worked for me.
Running the program in incognito tab will not cause this error. No need to clear caches also. See https://support.google.com/chrome/answer/95464?co=GENIE.Platform%3DAndroid&hl=en
I have pip-installed Flask and HTML5 on my Window-system. When I start the Hello World!-program with IDLE, I get a red message in the Python-Shell:
"* Running on xxxx://127.0.0.1:5000/". (xxxx = http)
And when I start it with app.run(debug=True) another red message appears:
"* Restarting with reloader".
My browser (Firefox) shows no reaction.
What can I do to get 'Hello World' in a new tab of Firefox?
The Code is:
from flask import Flask
app = Flask(__name__)
#app.route("/")
def hello():
return "Hello World!"
if __name__ == "__main__":
app.run(debug=True)
return and app.run are indended
You have to open a new tab with this url:
http://127.0.0.1:5000/
You need to actually open the page in your browser - it won't open itself. Open Firefox and navigate to
127.0.0.1:5000
(it's a URL)
When you run your code, it sits around waiting for a request from the user. When it gets a request, it'll return a response, and that's (sort of) what you see in your browser. Going to a URL is how you send that request - Flask will interpret anything sent to 127.0.0.1:5000 as a request, and try to match the URL to one of your #app.route decorators. For example, if you were to have a function decorated with #app.route("/hello"), then when you go to 127.0.0.1:5000/hello, Flask would run that function to determine the response.
Try out this code:
from flask import Flask
app = Flask(__name__)
#app.route("/")
def index():
return "<h1>Hello!</h1>"
if __name__ == "__main__":
from waitress import serve
serve(app, host="0.0.0.0", port=8080)
refrence Flask at first run: Do not use the development server in a production environment
from flask import Flask
app = Flask(__name__)
#app.route("/")
def hello():
return "Hello World!"
if __name__ == "__main__":
app.run(debug=True)
Try this, this works for me. Open your firefox browser and go to the address given in the output. ex: http://XXXX.X.X.X:5000/