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raise ValueError("cannot have a multithreaded and multi process server.") ValueError: cannot have a multithreaded and multi process server

Serving Flask app "server" (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: on
Running on http://0.0.0.0:9002/ (Press CTRL+C to quit)
Restarting with stat
Debugger is active!
Debugger PIN: 314-390-242
Exception in thread Thread-1:
Traceback (most recent call last):
File "/home/user/Downloads/yes/envs/emotionservice/lib/python2.7/threading.py", line 801, in __bootstrap_inner
self.run()
File "/home/user/Downloads/yes/envs/emotionservice/lib/python2.7/threading.py", line 754, in run
self.__target(*self.__args, **self.__kwargs)
File "/home/user/Downloads/yes/envs/emotionservice/lib/python2.7/site-packages/werkzeug/serving.py", line 963, in inner
fd=fd,
File "/home/user/Downloads/yes/envs/emotionservice/lib/python2.7/site-packages/werkzeug/serving.py", line 803, in make_server
raise ValueError("cannot have a multithreaded and multi process server.")
ValueError: cannot have a multithreaded and multi process server.

it solved.
set threaded=True
processes=1

If you want to use flask in multiprocessing mode (processes > 1). include threaded=False, otherwise it throws same error.
app.run(host=addr, port=port, threaded=False, processes=50)

how to make run Flask in Pythonista iOS?

I am using Pythonista app on iOS and I cant make Flask module work even with the most basic code.
my code is :
from flask import Flask
Flaskapp = Flask(__name__)
#Flaskapp.route('/')
def helloWorld():
return 'Woala'
Flaskapp.run(debug=True)
but keep receiving:
* Serving Flask app "app" (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: on
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
* Restarting with stat
Traceback (most recent call last):
File "/private/var/mobile/Library/Mobile Documents/iCloud~com~omz-software~Pythonista3/Documents/app.py", line 9, in <module>
Flaskapp.run(debug=True)
File "/private/var/mobile/Containers/Shared/AppGroup/C4F03A05-D634-4F19-865F-6072E8D6DC38/Pythonista3/Documents/site-packages-3/flask/app.py", line 990, in run
run_simple(host, port, self, **options)
File "/private/var/mobile/Containers/Shared/AppGroup/C4F03A05-D634-4F19-865F-6072E8D6DC38/Pythonista3/Documents/site-packages-3/werkzeug/serving.py", line 1007, in run_simple
run_with_reloader(inner, extra_files, reloader_interval, reloader_type)
File "/private/var/mobile/Containers/Shared/AppGroup/C4F03A05-D634-4F19-865F-6072E8D6DC38/Pythonista3/Documents/site-packages-3/werkzeug/_reloader.py", line 332, in run_with_reloader
sys.exit(reloader.restart_with_reloader())
File "/private/var/mobile/Containers/Shared/AppGroup/C4F03A05-D634-4F19-865F-6072E8D6DC38/Pythonista3/Documents/site-packages-3/werkzeug/_reloader.py", line 176, in restart_with_reloader
exit_code = subprocess.call(args, env=new_environ, close_fds=False)
File "/var/containers/Bundle/Application/E373BC34-F149-4EF5-9D0E-5CD807AB0771/Pythonista3.app/Frameworks/Py3Kit.framework/pylib/subprocess.py", line 268, in call
with Popen(*popenargs, **kwargs) as p:
File "/var/containers/Bundle/Application/E373BC34-F149-4EF5-9D0E-5CD807AB0771/Pythonista3.app/Frameworks/Py3Kit.framework/pylib/subprocess.py", line 708, in __init__
restore_signals, start_new_session)
File "/var/containers/Bundle/Application/E373BC34-F149-4EF5-9D0E-5CD807AB0771/Pythonista3.app/Frameworks/Py3Kit.framework/pylib/subprocess.py", line 1261, in _execute_child
restore_signals, start_new_session, preexec_fn)
PermissionError: [Errno 1] Operation not permitted
I have already installed flask trough pip in launch_stash.py

I received solution on another forum by user augusto, so I believe its worth sharing it here as well, as it works perfectly.
“The only way to use debug is:
if __name__ == '__main__':
app.run(use_reloader=False, debug=True)”
https://forum.omz-software.com/topic/5758/pythonista-flask-error-errno-1

The iOS sandbox has limitations that will prevent you from creating subprocesses
The flask development server is attempting to create a subprocess for detecting changes to your code that will then reload the server when you make changes. You can disable this by passing "--no-reload" flask run --no-reload. This would mean that you would have to manually restart your development server when you make changes

Change Pythonista’s default interpreter to Python 2.7.

Python flask reloader "Errno 2 No such file or directory" error when running on docker

I have a problem running a little Flask app with Docker. Everything works fine with debug off, but when I try to run it with debug on it fails at the reloader stage. It works fine on Windows, the problem only comes up with Docker.
In my docker repo I install caffe and all the dependencies I need (flask and wekrzeug). I thought the problem could be with the reloader so I also installed watchdog but it still shows up. Doesn't matter whether I run app.py itself or call python -m flask run.
docker run --volume=%cd%:/workspace -p 5001:5000 caffe:cpu
* Serving Flask app "app" (lazy loading)
* Environment: production
WARNING: Do not use the development server in a production environment.
Use a production WSGI server instead.
* Debug mode: on
* Running on http://0.0.0.0:5000/ (Press CTRL+C to quit)
* Restarting with inotify reloader
Traceback (most recent call last):
File "app.py", line 53, in <module>
app.run(debug = True, host='0.0.0.0')
File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 943, in run
run_simple(host, port, self, **options)
File "/usr/local/lib/python2.7/dist-packages/werkzeug/serving.py", line 988, in run_simple
run_with_reloader(inner, extra_files, reloader_interval, reloader_type)
File "/usr/local/lib/python2.7/dist-packages/werkzeug/_reloader.py", line 332, in run_with_reloader
sys.exit(reloader.restart_with_reloader())
File "/usr/local/lib/python2.7/dist-packages/werkzeug/_reloader.py", line 176, in restart_with_reloader
exit_code = subprocess.call(args, env=new_environ, close_fds=False)
File "/usr/lib/python2.7/subprocess.py", line 523, in call
return Popen(*popenargs, **kwargs).wait()
File "/usr/lib/python2.7/subprocess.py", line 711, in __init__
errread, errwrite)
File "/usr/lib/python2.7/subprocess.py", line 1343, in _execute_child
raise child_exception
OSError: [Errno 2] No such file or directory

I've encountered the same problem here and the solution for me was to remove the executable flag from the "run.py" file.
$ chmod -x run.py
The error was very similar, but in my case I'm using an Ubuntu 18.04 WSL on a Windows 10 machine.

Try to run the app as a module:
python -m flask run
See https://github.com/pallets/flask/issues/1829 for more information.

Flask Google Cloud App Engine: OSError: [Errno 98] Address already in use

I am trying to deploy a flask app on google cloud app engine. It runs smooth in my virtual environment locally but I get an 502 error running it in the cloud.
Now I am trying to debug my code on the cloud server, using debug mode and SSH into my instance. Using docker exec -it [ID] /bin/bash I am able to to access the root of my application. Now I upon running python app.py I get the following error:
* Serving Flask app "app" (lazy loading)
* Environment: production
WARNING: Do not use the development server in a production environment.
Use a production WSGI server instead.
* Debug mode: off
Traceback (most recent call last):
File "app.py", line 479, in <module>
app.run(port=8080)
File "/usr/local/lib/python3.6/site-packages/flask/app.py", line 941, in run
run_simple(host, port, self, **options)
File "/usr/local/lib/python3.6/site-packages/werkzeug/serving.py", line 814, in run_simple
inner()
File "/usr/local/lib/python3.6/site-packages/werkzeug/serving.py", line 774, in inner
fd=fd)
File "/usr/local/lib/python3.6/site-packages/werkzeug/serving.py", line 660, in make_server
passthrough_errors, ssl_context, fd=fd)
File "/usr/local/lib/python3.6/site-packages/werkzeug/serving.py", line 577, in __init__
self.address_family), handler)
File "/usr/local/lib/python3.6/socketserver.py", line 453, in __init__
self.server_bind()
File "/usr/local/lib/python3.6/http/server.py", line 136, in server_bind
socketserver.TCPServer.server_bind(self)
File "/usr/local/lib/python3.6/socketserver.py", line 467, in server_bind
self.socket.bind(self.server_address)
OSError: [Errno 98] Address already in use
I've been trying to kill those processes listed when I run:
ps -fA | grep python
However, this does not solve the problem of the address being in use. Also changing the port in the app.run() does not solve the issue for me.

I had a similar problem, it was caused by the Flask app also being run when the module was loaded, because I had
if __name__ == "__main__":
app.run()
at the bottom. Note that the recent requirement to name your sever file "main.py" could cause this bug to emerge.

I think the problem is that you don't need to specialize the port for cloud. Google Cloud finds the port to run your app on its own. So instead of app.run(port=8080) just write app.run()

While I was not able to figure out how to "free" the running address, I solved the problem by starting another flask process by running it on a different port like so:
flask run --port=80

Start HTTP/HTTPS server, python -m SimpleHTTPServer

I am trying to start HTTPS server on my localhost, but getting below error. I have opened cmd as an administrator.
C:\Windows\system32>python -m SimpleHTTPServer
Traceback (most recent call last):
File "C:\Python27\lib\runpy.py", line 162, in _run_module_as_main
"__main__", fname, loader, pkg_name)
File "C:\Python27\lib\runpy.py", line 72, in _run_code
exec code in run_globals
File "C:\Python27\lib\SimpleHTTPServer.py", line 224, in <module>
test()
File "C:\Python27\lib\SimpleHTTPServer.py", line 220, in test
BaseHTTPServer.test(HandlerClass, ServerClass)
File "C:\Python27\lib\BaseHTTPServer.py", line 595, in test
httpd = ServerClass(server_address, HandlerClass)
File "C:\Python27\lib\SocketServer.py", line 419, in __init__
self.server_bind()
File "C:\Python27\lib\BaseHTTPServer.py", line 108, in server_bind
SocketServer.TCPServer.server_bind(self)
File "C:\Python27\lib\SocketServer.py", line 430, in server_bind
self.socket.bind(self.server_address)
File "C:\Python27\lib\socket.py", line 224, in meth
return getattr(self._sock,name)(*args)
socket.error: [Errno 10013] An attempt was made to access a socket in a way forb
idden by its access permissions

Your windows firewall may block python from listening.
Additionaly, try adding a different port number higher than 1024 to the command line and see if it helps.

I was not able to start since some service is already running on 8000,
I was able to successfully start the service on another port 2000
C:\>python -m SimpleHTTPServer 2000
Serving HTTP on 0.0.0.0 port 2000 ..
One can get the PID/status for port 8000 by netstat -ano
C:\>netstat -ano
Active Connections
Proto Local Address Foreign Address State PID
TCP 0.0.0.0:912 0.0.0.0:0 LISTENING 2544
TCP 0.0.0.0:8000 0.0.0.0:0 LISTENING 4
TCP 0.0.0.0:19781 0.0.0.0:0 LISTENING 3100
shows PID 4 which is a system process.