I am trying to make a simple flask app using putty but it is not working
here is my hello.py file:
from flask import Flask
app = Flask(__name__)
#app.route('/')
def hello():
return 'Hello, World'
command I am running in putty (when in the file directory of hello.py)
pip install flask
python -c "import flask; print(flask.version)"
Output:1.1.2
export FLASK_APP=hello
export FLASK_ENV=development
flask run
when I go to the ip address: http://127.0.0.1:5000/
I get this
enter image description here
also here is the link of the instruction I am following/did:
https://www.digitalocean.com/community/tutorials/how-to-make-a-web-application-using-flask-in-python-3
if someone could let me know how to make it work would be great!
thank you!
I used my terminal on my window machine instead of putty which is a remote machine
Check your wlan0 inet adress by typing ifconfig in bash and use it, also try to add the following code in your flask app:
if __name__ == '__main__':
app.run(debug=True, port=5000, host='wlan0 IP Add')
Run the application by typing >>sudo python3 hello.py
Did this work?
I'm new to Flask. To launch the flask app, I did python -m flask run but I repeatedly get the error:
Failed to find Flask application or factory in module "app". Use "FLASK_APP=app:name to specify one.
I am using a virtualenv on a Windows 10 machine
from flask import Flask
app = Flask(__name__)
#app.route('/')
def index():
return "Hello, World!"
if __name__ == '__main__':
app.run(debug=True)
I expected a web server to start and to be able to navigate to localhost http://127.0.0.1:5000/ where I could view Hello, World! on my browser, but that never happened.
Instead of just "flask" use FLASK_APP=theflaskapp.py, like what Marco suggested:
env FLASK_APP=theflaskapp.py python -m flask run
This should fix it, if not, make sure you are executing the command to run the script in the same directory as it. You should also check if the problem is in flask or not by running "python theflaskapp.py" (In the same directory as the flask app still) and see if it works at all.
Reproducing the problem, and fixing it
Hello, I have reproduced the problem
This is the error code:
Error: Failed to find Flask application or factory in module
"src.app". Use "FLASK_APP=src.app:name to specify one.
Steps of reproducing the error:
Create a file called app.py
inside app.py put this code:
from flask import Flask
def create_app():
app = Flask("abc")
#app.route('/')
def hello_world():
return 'Hello, World!'
Or let the file be empty, like this:
# This is an empty file
# There is no flask application here
Inside CLI run these commands:
export FLASK_APP=app.py
flask run
watch the error appear on the screen
Solution 1 :
make the function return the app
Clearly create a variable called app and make it equal to the return value of the function, like this:
from flask import Flask
def create_app():
app = Flask(__name__)
#app.route('/')
def hello_world():
return 'Hello, World!'
return app
app = create_app()
Solution 2: Get the app outside of the function:
from flask import Flask
app = Flask("abc")
#app.route('/')
def hello_world():
return 'Hello, World!'
solution for flask Error: Could not locate a Flask application. You did not provide the "FLASK_APP" environment variable, and a "wsgi.py" or "app.py" module was not found in the current directory.
on windows when you tried set FLASKAPP=appname.py if your using PowerShell and if get the above error try this 👇👇
$env:FLASK_APP = "filename"
$env:FLASK_ENV = "development"
Flask run
try it if your having trouble launching the flask app.
You need to specify what file you want to work with. Try this.
For Windows:
set FLASK_APP=app.py
python -m flask run
For Mac OS and Linux:
export FLASK_APP=app.py
python -m flask run
I had a similar problem. It seems to work fine from the command prompt with python app.py. However, it would fail to launch with VS Code Debug.
The trick was to add:
server = app.server
before calling the app.run_server() portion
super simple in my case, and almost stupid...I had forgotten to save the file (CTRL+S), and I got exactly the same error. Once I saved the file, it worked directly!
Instead of providing a super straightforward "try this" suggestion, I'd recommend going straight to the flask source code to where this exception gets thrown, and reverse engineer from there. https://github.com/pallets/flask/blob/main/src/flask/cli.py#L78-L82
You can open this in your editor by going to the site-packages/flask/cli.py file and setting breakpoints around where this error gets thrown, and poke around from there.
Another working option is executing this command instead of flask run :
flask run --host=0.0.0.0
I had the same issue and I tried all the mentioned answers here but couldn't work finally this did the trick.
I'm getting the same error. The problem is you have to set the environment variable FLASK_APP in current project.
However, I have easiest way to solve this problem. Instead of using flask run command you can use python app.py.
"app.py" is your application entry point.
The simplest way to run your app without getting any issue is to create the app then run python app.py
from flask import (
Flask,
jsonify
)
# Function that create the app
def create_app(test_config=None ):
# create and configure the app
app = Flask(__name__)
# Simple route
#app.route('/')
def hello_world():
return jsonify({
"status": "success",
"message": "Hello World!"
})
return app # do not foget to return the app
APP = create_app()
if __name__ == '__main__':
# APP.run(host='0.0.0.0', port=5000, debug=True)
APP.run(debug=True)
On venv and my Windows10 only this syntax works:
set FLASK_APP=name_of_file:name_of_var_app
If file is application.py and 'app = Flask(__name__)',
run:
set FLASK_APP=application:app
You should change the name of the file, I have the same problem. So, I changed the name of the app file and It works for me. At first my app file name is 'socket.py' and I change it to 'app.py'.
My problem was that the python file didn't have execute permissions:
chmod a+x app.py
Similar issue! I was running the code in vs code. Installed python extension and after selecting a different python interpreter( in my case 3.10.5 64 bit), got a play button on the top right corner of my vs code interface.
That solved the issue for me.
In my case was a very simple issue, the activate.bat file was setup as Macintosh (CR) I changed to Windows (CR LF) and that worked for me.
Explanation: In windows OS, for the instruction to take effect, for example [set FLASK_ENV = development] a strong carriage return is needed, such as the (CR LF), that is what I could verify in my case. You can try to run the SET instruction in the console, cmd, and check if the FLASK settings are reflected. I understand that it is more a problem of the configuration of the environment and not of the application itself.
Try removing the spaces from this line of code, this is what fixed it for me when I had this error:
app=Flask(__name__)
I am new in python. I am trying to deploy python code on apache server for i.e i have created flask api. So for apache i have installed XAMPP and changed my httpd.conf to execute python on apache. It works well!! Here is code example which is working
Code working:
#!C:\Users\test.lab\AppData\Local\Continuum\anaconda3\envs\myproject\python.exe
# enable debugging
print("Content-type: text/html\n")
print ("Hello Python Web Browser!! This is cool!!")
But when I tried to import that through 500 Error, Here is the code
#!C:\Users\test.lab\AppData\Local\Continuum\anaconda3\envs\myproject\python.exe
# enable debugging
from flask import Flask
app = Flask(__name__)
#app.route('/')
def index():
return 'testing'
if __name__ == '__main__':
app.run(debug = True)
flask is installed on my environment (myproject). When I run through command like python test.py and it works.
Flask has it's own development web server.
Using python myfile.py which will work properly as a webserver (no need for apache on development).
If you still want do deploy on Apache, Flask have some info on how to do so, docs: http://flask.pocoo.org/docs/1.0/deploying/mod_wsgi/
Special attention to this: http://flask.pocoo.org/docs/1.0/deploying/mod_wsgi/#creating-a-wsgi-file
In the flask doco the following description is shown of deploying a flask app under twistd.
twistd web --wsgi myproject.app
I have a foo.py which looks like this
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=8080)
So I expected to be able to run that under twistd like this
twistd web --wsgi foo.app
but twistd doesn't like that (just spits out the help text).
What am I doing wrong ?
BTW in case it matters I'm running this in a virtualenv (in which I have installed both flask and twisted) and the current directory when I issue the twistd command contains foo.py .
EDIT: The version of twistd I am using is 18.7.0
I had failed to notice (until prompted to by Peter Gibson's comment ) that after the help text appears the message "No such WSGI application: 'foo.app'" appears.
You need to add the current directory to the PYTHONPATH environment variable. Try
PYTHONPATH=. twistd web --wsgi foo.app
Or on Windows (untested)
set PYTHONPATH=.
twistd web --wsgi foo.app
I have simple Flask app test.py:
from flask import Flask
app = Flask(__name__)
#app.route('/')
def test():
return 'Hello world!'
if __name__ == '__main__':
app.run()
Run over twisted 16.3.0 work fine:
twistd -n web --port 5000 --wsgi test.app
After upgrade twisted to 16.4.0 i have got error on start:
No such WSGI application: 'test.app'
What is mean?
Your are likely picking up the test module which is part of the Python standard library. Rename your code file (module) to something else. You also may need to set PYTHONPATH so it looks in the directory where code module is.