I am trying to learn Flask using VScode.
The tutorial that I am following is: Python Flask Tutorial: Full-Featured Web App Part 1 - Getting Started.
I did the following things:
Created a new virtualenv in a folder using: virtualenv venv
activated it as: venv\Scripts\activate (I am on Windows 10)
After that, I created a new directory named Flask_Blog using mkdir Flask_Blog and in it, I created a new flaskblog.py file containing the following code:
from flask import Flask
app = Flask(__name__)
#app.route('/')
def hello():
return 'Hello'
Then, in the terminal of VScode, I changed my working directory in order to be in the Flask_Blog directory using cd Flask_Blog.
Now, when I am doing set FLASK_APP=flaskblog.py followed by flask run, I am getting the following error:
(venv) PS C:\Users\kashy\OneDrive\Desktop\Flask\Flask_Blog> flask run
* Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
Use a production WSGI server instead.
* Debug mode: off
Usage: flask run [OPTIONS]
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.
But
When I do the same in the cmd prompt, the code runs and I get to see the output.
I am completely new to this. Can anyone please tell me what is the mistake I am doing in VSCode and why is it working in the cmd?
Issue raised in VsCode
Under Powershell, you have to set the FLASK_APP environment variable as follows:
$env:FLASK_APP = "webapp"
Then you should be able to run "python -m flask run" inside the hello_app folder. In other words, PowerShell manages environment variables differently, so the standard command-line "set FLASK_APP=webapp" won't work.
Try Set FLASK_APP = Full path of the folder/filename.py.
This worked for me
This worked for me on the VSCode:
$env:FLASK_APP= 'C:\Python\ex003\main:app'
Related
I am trying to learn flask. I created a venv and installed flask. I set the flaskblog.py as FLASK_APP. But it says serving flask app 'app.py' and it fails. why is this?
I'm not sure if the image is visible. So, I'll write what I've tried below,
In the windows powershell, I cd to the directory where my venv and python script is, then
$ . vnev\Scripts\activate
$ pip install flask
$ set FLASK_APP=flaskblog.py
$ flask run
it gives me the error saying,
Serving Flask app 'app.py' (lazy loading)
Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
Use a production WSGI server instead.
Debug mode: off
Usage: flask run [OPTIONS]
Try 'flask run --help' for help.
Error: Could not import 'app'.
What is wrong here? I have also tried $ setx FLASK_APP "flaskblog.py and still get the same result. I think i created this app.py using pycharm's virtual environment and now, I cannot set other files to FLASK_APP.
What should I do?
I tried everything but couldn't import app.py. It gives me the following error
set FLASK_APP = "C:\Users\Hp\PycharmProjects\sakshi.py\app.py"
(env) C:\Users\Hp\Documents\flask_app>flask run
Serving Flask app " app.py"
Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
Use a production WSGI server instead.
Debug mode: off
Usage: flask run [OPTIONS]
Error: Could not import " app".
Your help would be much appreciated!
Change the sever environment to development env.
export FLASK_APP=yourapp
export FLASK_ENV=development
and then start the server with:
flask run
If you still get error try this way:
set FLASK_ENV=development
flask run
And then run your python file.
At least one problem: spaces in the set command. From the syntax in the microsoft doc:
set [<Variable>=[<String>]]
set [/p] <Variable>=[<PromptString>]
set /a <Variable>=<Expression>
There is no space on either side of the = in the set command.
And, the path name "C:\Users\Hp\PycharmProjects\sakshi.py\app.py" is questionable. It is, at the very least, a bad idea to name a directory sakshi.py.
I could see the problem is with space between the variable and value in set command,
In windows,
set var=value
In *nix family,
export var=value
Your script name is app.py which by default flask recognizes as entry point for the app.
You only have to explicitly set the FLASK_APP variable when you are running from another directory (which you are) or the name of the script is something other than app.py. You can directly run the flask run command from the project base directory itself.
Here is what I did:
From start, opened Anaconda prompt
Did a conda create -n flaskEnv pip flask
Did a conda activate flaskEnv
Did a cd to my desired folder
Created a file named flaskblog.py with following code:
from flask import Flask
app = Flask(__name__)
#app.route("/")
def hello():
return "<h1>Hello World</h1>"
Did a set FLASK_APP=flaskblog.py and then set FLASK_ENV=development
Now, I do flask run but it throws an error 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.
However, when I add the following code on my file flaskblog.py and then do a python flaskblog.py
, it runs fine and serves on my localhost.
if __name__=='__main__':
app.run(debug=True)
I am just trying to understand what am I doing wrong when doing a flask run from anaconda prompt.
Thanks
I had the same problem and I just closed the terminal and switched everything relating to the project off and then tried again and it worked just fine.
I don't understand why if I want to run Flask application I need
(venv) $ export FLASK_APP=microblog.py
(venv) $ flask run
But if I want to run Django application I only
(venv) $ python manage.py runserver
without export DJANGO_APP=microblog.py
Why? Why I need the export app in the first case, but in the second case I don't need?
Firstly, Django and Flask are different frameworks. There's no reason why the commands to start them should be the same.
You need to export FLASK_APP to tell flask which app to run.
Doing export FLASK_APP=microblog.py sets an environment variable FLASK_APP. The flask application can then read this variable from the environment and use it to run the application.
In Python you can access environment variables from os.environ, or use the os.getenv method:
import os
flask_app = os.getenv('FLASK_APP')
If you use the django-admin command, you need to export DJANGO_SETTINGS_MODULE in a similar way:
$ export DJANGO_SETTINGS_MODULE=yourproject.settings
$ django-admin runserver
However with Django, you usually use runserver with manage.py instead of django-admin. The manage.py is specific to your project and sets the DJANGO_SETTINGS_MODULE environment variable if it hasn't been set already:
os.environ.setdefault("DJANGO_SETTINGS_MODULE", "yourproject.settings")
Therefore you don't need to export DJANGO_SETTINGS_MODULE when using manage.py.
From a source code prospective, seems that the FLASK_APP variabile is used for understand which flask app run in a 'multi-flask-app' environment.
Is necessary only if the app_name from the source is not present.
The FLASK_APP env variable is used only into the find_best_app method of the cli.py file of the flask framework.
Simply in your project, you don't need something like export FLASK_APP=microblog.py unless you want environment variables
from flask import jsonify,Flask
app = Flask(__name__)
#app.route("/<Your Route>/<string:<Your Param>>")
def main(<Your Param>):
//DO LOGIC HERE
data =[{'TestData1' : "" ,<YOUR OUTPUT>}]
return jsonify(data), 200
app.run(debug=False,host="0.0.0.0",port=<PORT YOU WANT TO HOST>)
Problem: When you write the command flask run in the console, how does flask know which file to run?
Solution: Thats why we use export FLASK_APP=microblog.py
It is setting the FLASK_APP (an internal flask variable) value to microblog.py
It tells flask to use microblog.py as start-up file for the application when you run flask via flask run command.
If you decide not to do so, then when you run flask run, then there is no way for flask to know which file to run. Now you could run the application using python filename.py instead of flask run
So python microblog.py in your case.
I keep getting this error flask.cli.NoAppException: The file/path provided (new_app.py) does not appear to exist. Please verify the path is correct. If app is not on PYTHONPATH, ensure the extension is .py it goes away after I restart the Flask server.
I am running flask run in the correct directory where my app is. This just started happening after working for 2 weeks. I've read that it could be due to an import error, but I am not finding any modules that are not installed on my virutalenv.
from flask import Flask
app = Flask(__name__)
app.debug=True
Most likely you haven't set the FLASK_APP environment variable.
To run the application you can either use the flask command or
python’s -m switch with Flask. Before you can do that you need to tell
your terminal the application to work with by exporting the FLASK_APP
environment variable:
$ export FLASK_APP=hello.py
$ flask run * Running on http://127.0.0.1:5000/
If you are on Windows you need to use set
instead of export.
Alternatively you can use python -m flask:
$ export FLASK_APP=hello.py
$ python -m flask run * Running on http://127.0.0.1:5000/
EDIT
If you have FLASK_APP set then try adding this to new_app.py
app.run(debug=True, port=8800)
Or if you're on Windows:
if __name__ == '__main__':
app.run(debug=True, port=8800)
And then just execute the app with python new_app.py.