I just started learning flask and I am stuck at setting up the Flask environment variables. I don't know how to setup the environment variables. Whenever I use the flask run command, I encounter the following error:
Error message : 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.
I did a lot of google searches to setup environment variables on Windows but I am unable to find a solution and sometimes I am not able to understand the solution.
How to do this ? How to get the "app.py" or "wsgi.py" ?
Windows PowerShell
set FLASK_APP=hello.py
$env:FLASK_APP = "hello.py"
flask run
you need to provide an application environment. So Flask needs to know, the .py file to run.
Try to run that command
export FLASK_APP=application.py
where application.py - name of your Flask app in my case it is application.py.
after that
flask run
I used this and it worked because I am doing it in Windows PowerShell.
$env:FLASK_APP = "app.py"
however, flask run didn't and gave me could not import application.
My error was also same but fortunately I was able to resolve it. Here you go,
D:\Development\Projects\Python_Projects\flask_blog>set FLASK_APP=app.py
D:\Development\Projects\Python_Projects\flask_blog>$env:FLASK_APP = "app.py"
D:\Development\Projects\Python_Projects\flask_blog>python -m flask run
hope this could help someone,
first set flask env like this inside the python virtual env
(for windows command prompt)
set FLASK_ENV=development
then
set FLASK_APP=app.py
(app.py should be your flask application file)
I don't think the 'flask run' command is the one which causes the error here.
I got the same message error and the problem came from the fact I copied/pasted the set FLASK_APP and $env: FLASK_APP commands as written in the documentation. I had to add spaces before and after '>' or '=', and then everything worked.
Example: this command didn't work 'C:\path\to\app>set FLASK_APP=hello.py', but this one did 'C:\path\to\app > set FLASK_APP = hello.py'.
Maybe it's the same problem you have?
You need to actually run it from your Windows command line, NOT the built in command line for something like Visual Studio Code. Run those commands from your windows command line, in the proper directory, and everything should work.
The reason this creates problems - Visual Studio Code creates a Powershell environment for your command line. You could use the recommended $env:FLASK_APP = "your_app.py" from within the VSC environment and that will work too.
A bit late but I hope this helps others!!!
I was facing the same problem and the thing that worked for me was don't put spaces before and after the = sign.
For example: FLASK_APP = flaskblog.py is wrong and will likely give you an error because of the spaces before and after the = sign.
Instead try FLASK_APP=flaskblog.py
It worked for me.
You can try this
set FLASK_APP=application.py
flask run
If you're using powershell, make sure you add quotations when setting the environment variable:
$env:FLASK_APP = "app.py
Then flask run should work.
A step-wise solution is provided below:
Go to the folder where you have placed your flask app (on the command line)
Create a virtual environment as using the command ($ py -m venv env) here 'venv' is the short form of the virtual environment and 'env' at the end represents the name of the environment which you want (I have named it as env). Thereafter you can see at from the file explorer that a folder named 'env' is created in the folder stated at point #1 above.
Enter the following command ($env\Scripts\activate) by pressing enter this will turn on your virtual environment
Thereafter, enter the following command ($set FLASK_APP=<your app name>.py)
Enter the following command ($flask run)
The set command works but to setup the environment, you need to make sure that you are in the right directory where the file is located. For example, if my application is hello_world.py and it is located at the venv\hello\hello_world.py, you need to make sure that you are in the right directory before setting up set FLASK_APP=hello_world.py this is for windows but in another OS, you need to use export instead set
(venv) C:\Users\myProjects\venv\hello\set FLASK_APP=hello_world.py
You're typing in the commands that look correct. It may be a Windows security item that your user cannot make changes to environmental variables. If you're on Windows 10, search "View Advanced System Settings." Then click environmental variables, hit new user variable and make it FLASK_APP and set the path where it asks. Then do flask run in terminal.
To add: if your terminal is just not cooperating with your environment variables, you can just call the method inside the script itself, avoiding this error completely.
from flask import Flask
app = Flask(__name__)
#your decorators, etc here
if __name__=='__main__':
app.run()
Then when you're in your activated virtual environment, all that's needed is
$python flaskblog.py
I tried a few ways mentioned in this thread so thanks everyone for your inputs.This is what worked for me.
1st, making sure you're running it when you're on the main .py file;
2nd, if you're on a Windows, use set FLASK_APP="your_file.py" (don't forget the quotation marks)
and then flask db init to initialize
...woo so happy my app finally worked!
I resolved similar problem by running the Command Prompt with admin rights (Windows+R --> enter cmd --> Hold the keys Ctrl+Shift+Enter) and then running the below command:
set FLASK_APP=<ProgramName>.py
The similar error was appearing when I was trying to run the application. I have to change the path. I changed directory to the folder where hello.py was saved.
(venv) C:\Users\win10\flask-tutorial\myproject>cd venv
(venv) C:\Users\win10\flask-tutorial\myproject\venv>set FLASK_APP=hello.py
(venv) C:\Users\win10\flask-tutorial\myproject\venv>flask run
You need to specify the environment of the application. Like this
export FLASK_APP = application.py
after performing this operation
flask run
But my suggestion is that it will be easier for you to perform these operations there while creating your application, rather than constantly stating this in the terminal. After reviewing the documentation, if you do what I said, it will be enough to come to the terminal and run python app.py runserver in terminal.
You can check flask's documentation for that.
https://flask.palletsprojects.com/en/1.1.x/patterns/appfactories/
I just had the same problem on Windows, using the command line. And problem was that I wasn't setting my app as the one that flask needs to run.
So, in my case, I turned the debug mode on first (not required but convenient) with:
set FLASK_ENV=development
then:
set FLASK_ENV=server.py
#where server.py is the name of my "app"
and finally ran flask:
flask run
Remove space between FLASK_APP and flaskblog.py, then execute flask run command or just run the program like other program python flaskblog.py
I had the same issue. These steps helped me:
Delete existing venv environment, then in terminal type in project folder path py -3 venv venv
Then: venv\Scripts\activate
pip install flask
set FLASK_APP=hello
flask run
I was facing the same issue when doing flask run in the VScode.
But when I tried on CMD it worked fine
DO one thing
from flask import Flask
app = Flask(__name__)
#app.route('/')
def hello_world():
return 'Hello World!'
app.run(debug=True)
python .\app.py
As Windows user, what helped me:
set FLASK_APP=appname.py | in CMD
$env:FLASK_APP="appname.py" | in PowerShell or VS Code(your code editing app)
And next line: flask run
for me adding this line at the end of the code fix it
if __name__=='__main__':
app.run()
so complete code would be this:
for more solution check this out
common error with running flask in windows
In the command line, you'll run three lines of code. The first two lines tell the terminal where to find your application and to run it in development mode, which allows you to keep it running while it hotloads any modifications. The third actually starts the application.
# For Mac/Linux
export FLASK_APP=app.py
export FLASK_ENV=development
flask run
# For Windows
set FLASK_APP=app.py
set FLASK_ENV=development
flask run
set FLASK_APP=app.py
python -m flask run
i use it and working hope helpful
I had a same problem but i solve the issue by pasting the code in the python file.
if __name__=="__main__":
app.run(debug=True)
Related
I have a flask project and specified a .flaskenv file like this:
FLASK_APP=my_program.py
FLASK_DEBUG=1
I'm running this on a MacOS system with Python 3.8.x.
I'm calling flask run to start the program. This works fine, but I want the program to run in a conda environment, however, no matter in which conda environment I'm using, the program will always run in the local python environment.
Just using python my_program.py will work and use the desired conda environment that is active in the shell.
But, if possible, I would like to specify the python environment that is used when calling flask run.
Is there some way to specify the desired python environment for example in .flaskenv?
Edit:
Installing flask via conda in the respective environment solved the problem for me. For further details see answer by #VPfB and comments.
As discussed in the comments, the flask Python script is not able to select the environment it will run inside, because it is a part of some environemnt.
The script contains startup code which imports the entry point function from the flask library (package) and invokes it. The script is part of the package and the package is installed in some Python environment (or system-wide).
In order to be able to run a flask app in multiple environments, it must be installed in each of them.
This is not the same as "What is the purpose of the -m switch" as I'm talking about a function that runs independently (flask) and wanting to know why it works differently with -m. The answers there do not resolve the question.
I have a flask app that I'm building. It's set up with a subdirectory for the code, so
setup.py
gallery/
__init.py__
models.py
view.py
When I run it using 'python -m flask run' it works. When I run it just as 'flask run' it doesn't. I haven't been able to find anyone who knows what the actual difference is between the two commands, so I'm kind of stuck.
Here's the output from 'flask run':
(venv)MacBook-Pro-4:Finished khunter$ export FLASK_APP=gallery
(venv)MacBook-Pro-4:Finished khunter$ flask run
Usage: flask run [OPTIONS]
Error: The file/path provided (gallery) does not appear to exist. Please
verify the path is correct. If app is not on PYTHONPATH, ensure the
extension is .py
Any thoughts about where the difference lies so I can use 'flask run' appropriately?
EDIT: It looks like I can get through this with pip install --editable to grab the setup.py I have there.
There isn't many differences. When you run python -m flask run, you are running flask/__main__.py.
While when you run flask run, you are running flask/cli.py.
As you can see, the first one is just an alias of the second one. Although it passes as_module == True. The reason why it need to be additionally processed is showed in comment. You can check it by yourself.
I believe your issue is caused by path problem. Double check if your path is correct.
I can run my app from the console that there is in pyCharm but If I try to run my app from a shell my app doesn't find "pymysql" module.
The module is installed in my project in a virtual environment. You can see in the next image how is installed this module.
And If try to run my app from the shell I've got this error:
I'm using python3.
What am I doing wrong? Is there any easy way to access to the module?
There are several ways:
activate virtual env: source venv/bin/activate.
directly use specific python: venv/bin/python main.py
Surely you can temporarily add venv/bin to your PATH, that's almost the same as the first option: export PATH=full/path/to/bin:$PATH
Generally I recommend the first option. But sometimes you may want to use the second one. For example, you want to use this python in a crontab script.
I'm building a Flask application and my file "helloworld.py" is:
from flask import Flask
app = Flask(__name__)
#app.route('/home')
def hello_world():
return 'Home!'
#app.route('/about')
def about_us():
return 'aboutus!'
My Flask code after activating venv:
set FLASK_DEBUG=1
flask run
Changes made to my "helloworld.py" file doesn't get automatically updated. I still have to restart it manually as I change anything in the code.
There is no error thrown but the changes simply don't reflect in the browser.
Why is the debug mode not working, can someone help me out?
The syntax for setting environment variables is different in PowerShell, use $env:.
> $env:FLASK_APP = "helloworld.py"
> $env:FLASK_DEBUG = "1"
> flask run
set is used for Windows CMD.
> set FLASK_DEBUG=1
export is used by most other shells, like Bash and Zsh, such as on Linux and MacOS.
$ export FLASK_DEBUG=1
If your OS is Windows, you can make use of VSC terminal an run the same code
set FLASK_DEBUG=1
flask run
Make sure you are only making changes inside py file, not in the terminal, otherwise it won't work.
if you are on Linux, use 'export' command instead of 'set'
and yes, its not working on power shell use CMD, but even this did not work properly.so I had to use embedded terminal. but the good thing is that you can use normal CMD(not PowerShell) as embedded terminal.
so the solution is, do not use PowerShell.
Solution to set FLASK_DEBUG=1 not working is,
Note- terminal->Use cmd terminal in VSCODE and NOT powershell
do the required on cmd terminal in vscode which is
set FLASK_APP=app.py
set FLASK_DEBUG=1
flask run
My computer says...
"-bash: appcfg.py: command not found"
What is wrong?
I can run my application using google-app-engine-launcher and I have python pre-installed.
I am trying to upload my app using "appcfg.py update myapp"
I am new to Mac development.
In App Engine launcher there is a menu option called "Make Symlinks..." that adds symlinks for the various App Engine utility commands, like appcfg.py.
This is how my path dir looks like: Home/Brice/google_projects/google_appengine
I store both the google_appengine and my google_apps in my google_projects folder
In terminal: (While in my google_projects folder)
upload to localhost:
google_appengine/dev_appserver.py appname
upload to GAE:
google_appengine/appcfg.py update appname
and replace appname with the name of your app folder
Hope that helps!
If someone (like me) comes across this more recently due to appcfg.py and dev_appserver.py still appearing frequently in the documentation:
0.9.68 (2015/07/08)
[...]
The standalone App Engine SDKs are no longer distributed through the Cloud
SDK.
App Engine functionality can still be used through the
gcloud preview app command group.
[...]
If you need to use appcfg or dev_appserver directly, these are still
available in the App Engine SDK downloads that can be found here:
https://cloud.google.com/appengine/downloads
(from google-cloud-sdk/RELEASE_NOTES)
Try: ./appcfg.py
Current dir is usually not part of path.
If is not in a directory specified in the PATH environment variable and marked executable it wont execute by calling its plain name.
when in doubt the following should always work:
python /path/to/appcfg.py <your arguments>
Because the top-voted and accepted answer doesn't explain this, and not everyone will read the comments on it, here's what to do:
Ensure you've installed Google App Engine SDK/Launcher from https://cloud.google.com/appengine/downloads?csw=1
Within it, select the option to "Make Symlinks...". "Make Command Symlinks?" may pop up in a dialog when you open it for the first time or after it's updated itself.
You'll have to do this each time it updates itself or it'll stop working. This is often what's gone wrong.
Using command line there are two options
1. make the two files executable and create symbolic links for them
# chmod +x path/to/google_appengine/dev_appserver.py
# ln -s /path/to/google_appengine/dev_appserver.py /bin
# chmod +x path/to/google_appengine/appcfg.py
# ln -s /path/to/google_appengine/appcfg.py /bin
2. export PATH and PYTHONPATH variables. To do this add following lines in .bashrc file
export PATH=$PATH:/path/to/google_appengine/
export PYTHONPATH="$PYTHONPATH:/path/to/google_appengine:/path/to/google_appengine/lib/:/path/to/google_appengine/lib/yaml/"