I want to import some function in form of a script, let's call it controller.py, to a flask app as a web service. Let's call the flask app api.py.
The problem is, in controller.py, there is a pyserial declaration.
controller.py:
import serial
ser = serial.Serial('COM31', 9600,timeout=2)
def serial_function(foo):
ser.write(foo)
reply = ser.read()
return reply
api.py:
from flask import Flask
import controller as cont
app = Flask(__name__)
#app.route('/function/<foo>',methods=['GET'])
def do_function(foo):
data=cont.serial_function(foo)
return data
if __name__ == '__main__':
app.run('0.0.0.0', 80,True)
But i got this error:
raise SerialException("could not open port %s: %s" % (self.portstr, ctypes.WinError()))
serial.serialutil.SerialException: could not open port COM31: [Error 5] Access is denied.
It seems that Flask is trying to import controller.py over and over again, and the serial port re initialized.
Is there some way I can achieve what I'm trying to do as described above?
The main problem of your code is that the Serial object creation is directly in the code of your module. This way, each time you import your module, python will load/interpret the file, and by doing this, will execute all the code found at the root level of your module. (you can refer to this excellent answer if you want to dig in more : What does if __name__ == "__main__": do?)
Moreover, in debug mode, Flask will start 2 processes (one to monitor the source code and when changed, restart the second one, which is the one who will really handle the requests) and in production mode, you could create much more threads or processes when starting the server, by doing this, your module is imported at least twice => conflict in serial open.
A possible solution would be to remove the initialization of the serial port from your module and use the context manager syntax in your method :
def serial_function(foo):
with serial.Serial('COM31', 9600,timeout=2) as ser:
ser.write(foo)
reply = ser.read()
return reply
This way, you will open (and close) your serial port at each read.
But, you'll still have to deal with concurrent access if you have multiple clients making requests to your webserver simultaneously.
EDIT:
If, as you say in comment, you need to open only once your serial port, you'll need to encapsulate this in a specific object (probably using a Singleton pattern) that will be responsible of opening the serial port if not already opened:
class SerialProxy:
def __init__(self):
self.__serial = None
def serial_function(self, foo):
if self.__serial is None:
self.__serial = serial.Serial('COM31', 9600,timeout=2)
self.__serial.write(foo)
reply = self.__serial.read()
return reply
I'm writing a small web server in Python, using BaseHTTPServer and a custom subclass of BaseHTTPServer.BaseHTTPRequestHandler. Is it possible to make this listen on more than one port?
What I'm doing now:
class MyRequestHandler(BaseHTTPServer.BaseHTTPRequestHandler):
def doGET
[...]
class ThreadingHTTPServer(ThreadingMixIn, HTTPServer):
pass
server = ThreadingHTTPServer(('localhost', 80), MyRequestHandler)
server.serve_forever()
Sure; just start two different servers on two different ports in two different threads that each use the same handler. Here's a complete, working example that I just wrote and tested. If you run this code then you'll be able to get a Hello World webpage at both http://localhost:1111/ and http://localhost:2222/
from threading import Thread
from SocketServer import ThreadingMixIn
from BaseHTTPServer import HTTPServer, BaseHTTPRequestHandler
class Handler(BaseHTTPRequestHandler):
def do_GET(self):
self.send_response(200)
self.send_header("Content-type", "text/plain")
self.end_headers()
self.wfile.write("Hello World!")
class ThreadingHTTPServer(ThreadingMixIn, HTTPServer):
daemon_threads = True
def serve_on_port(port):
server = ThreadingHTTPServer(("localhost",port), Handler)
server.serve_forever()
Thread(target=serve_on_port, args=[1111]).start()
serve_on_port(2222)
update:
This also works with Python 3 but three lines need to be slightly changed:
from socketserver import ThreadingMixIn
from http.server import HTTPServer, BaseHTTPRequestHandler
and
self.wfile.write(bytes("Hello World!", "utf-8"))
Not easily. You could have two ThreadingHTTPServer instances, write your own serve_forever() function (don't worry it's not a complicated function).
The existing function:
def serve_forever(self, poll_interval=0.5):
"""Handle one request at a time until shutdown.
Polls for shutdown every poll_interval seconds. Ignores
self.timeout. If you need to do periodic tasks, do them in
another thread.
"""
self.__serving = True
self.__is_shut_down.clear()
while self.__serving:
# XXX: Consider using another file descriptor or
# connecting to the socket to wake this up instead of
# polling. Polling reduces our responsiveness to a
# shutdown request and wastes cpu at all other times.
r, w, e = select.select([self], [], [], poll_interval)
if r:
self._handle_request_noblock()
self.__is_shut_down.set()
So our replacement would be something like:
def serve_forever(server1,server2):
while True:
r,w,e = select.select([server1,server2],[],[],0)
if server1 in r:
server1.handle_request()
if server2 in r:
server2.handle_request()
I would say that threading for something this simple is overkill. You're better off using some form of asynchronous programming.
Here is an example using Twisted:
from twisted.internet import reactor
from twisted.web import resource, server
class MyResource(resource.Resource):
isLeaf = True
def render_GET(self, request):
return 'gotten'
site = server.Site(MyResource())
reactor.listenTCP(8000, site)
reactor.listenTCP(8001, site)
reactor.run()
I also thinks it looks a lot cleaner to have each port be handled in the same way, instead of having the main thread handle one port and an additional thread handle the other. Arguably that can be fixed in the thread example, but then you're using three threads.
I'm implementing a bi-directional ping-pong demo app between an electron app and a python backend.
This is the code for the python part which causes the problems:
import sys
import zerorpc
import time
from multiprocessing import Process
def ping_response():
print("Sleeping")
time.sleep(5)
c = zerorpc.Client()
c.connect("tcp://127.0.0.1:4243")
print("sending pong")
c.pong()
class Api(object):
def echo(self, text):
"""echo any text"""
return text
def ping(self):
p = Process(target=ping_response, args=())
p.start()
print("got ping")
return
def parse_port():
port = 4242
try:
port = int(sys.argv[1])
except Exception as e:
pass
return '{}'.format(port)
def main():
addr = 'tcp://127.0.0.1:' + parse_port()
s = zerorpc.Server(Api())
s.bind(addr)
print('start running on {}'.format(addr))
s.run()
if __name__ == '__main__':
main()
Each time ping() is called from javascript side it will start a new process that simulates some work (sleeping for 5 seconds) and replies by calling pong on nodejs server to indicate work is done.
The issue is that the pong() request never gets to javascript side. If instead of spawning a new process I create a new thread using _thread and execute the same code in ping_response(), the pong request arrives in the javascript side. Also if I manually run the bash command zerorpc tcp://localhost:4243 pong I can see that the pong request is received by the nodejs script so the server on the javascript side works ok.
What happens with zerorpc client when I create a new process and it doesn't manage to send the request ?
Thank you.
EDIT
It seems it gets stuck in c.pong()
Try using gipc.start_process() from the gipc module (via pip) instead of multiprocessing.Process(). It creates a new gevent context which otherwise multiprocessing will accidentally inherit.
In my Tornado app in some situation some clients disconnect from server but my current code doesn't detect that client is disconnect from server. I currently use ping to find out if client is disconnected.
here is my ping pong code:
from threading import Timer
class SocketHandler(websocket.WebSocketHandler):
def __init__(self, application, request, **kwargs):
# some code here
Timer(5.0, self.do_ping).start()
def do_ping(self):
try:
self.ping_counter += 1
self.ping("")
if self.ping_counter > 2:
self.close()
Timer(60, self.do_ping).start()
except WebSocketClosedError:
pass
def on_pong(self, data):
self.ping_counter = 0
now I want to set SO_RCVTIMEO in tornado instead of using ping pong method.
something like this :
sock.setsockopt(socket.SO_RCVTIMEO)
Is it possible to set SO_RCVTIMEO in Tornado for close clients from server after specific time out ?
SO_RCVTIMEO does not do anything in an asynchronous framework like Tornado. You probably want to wrap your reads in tornado.gen.with_timeout. You'll still need to use pings to test the connection and make sure it is still working; if the connection is idle there are few guarantees about how long it will take for the system to notice. (TCP keepalives are a possibility, but these are not configurable on all platforms and generally use very long timeouts).
When starting a bottle webserver without a thread or a subprocess, there's no problem. To exit the bottle app -> CTRL + c.
In a thread, how can I programmatically stop the bottle web server ?
I didn't find a stop() method or something like that in the documentation. Is there a reason ?
For the default (WSGIRef) server, this is what I do (actually it is a cleaner approach of Vikram Pudi's suggestion):
from bottle import Bottle, ServerAdapter
class MyWSGIRefServer(ServerAdapter):
server = None
def run(self, handler):
from wsgiref.simple_server import make_server, WSGIRequestHandler
if self.quiet:
class QuietHandler(WSGIRequestHandler):
def log_request(*args, **kw): pass
self.options['handler_class'] = QuietHandler
self.server = make_server(self.host, self.port, handler, **self.options)
self.server.serve_forever()
def stop(self):
# self.server.server_close() <--- alternative but causes bad fd exception
self.server.shutdown()
app = Bottle()
#app.route('/')
def index():
return 'Hello world'
#app.route('/stop') # not working from here, it has to come from another thread
def stopit():
server.stop()
server = MyWSGIRefServer(port=80)
try:
app.run(server=server)
except:
print('Bye')
When I want to stop the bottle application, from another thread, I do the following:
server.stop()
I had trouble closing a bottle server from within a request as bottle seems to run requests in subprocesses.
I eventually found the solution was to do:
sys.stderr.close()
inside the request (that got passed up to the bottle server and axed it).
An updated version of mike's answer.
from bottlepy.bottle import WSGIRefServer, run
from threading import Thread
import time
class MyServer(WSGIRefServer):
def run(self, app): # pragma: no cover
from wsgiref.simple_server import WSGIRequestHandler, WSGIServer
from wsgiref.simple_server import make_server
import socket
class FixedHandler(WSGIRequestHandler):
def address_string(self): # Prevent reverse DNS lookups please.
return self.client_address[0]
def log_request(*args, **kw):
if not self.quiet:
return WSGIRequestHandler.log_request(*args, **kw)
handler_cls = self.options.get('handler_class', FixedHandler)
server_cls = self.options.get('server_class', WSGIServer)
if ':' in self.host: # Fix wsgiref for IPv6 addresses.
if getattr(server_cls, 'address_family') == socket.AF_INET:
class server_cls(server_cls):
address_family = socket.AF_INET6
srv = make_server(self.host, self.port, app, server_cls, handler_cls)
self.srv = srv ### THIS IS THE ONLY CHANGE TO THE ORIGINAL CLASS METHOD!
srv.serve_forever()
def shutdown(self): ### ADD SHUTDOWN METHOD.
self.srv.shutdown()
# self.server.server_close()
def begin():
run(server=server)
server = MyServer(host="localhost", port=8088)
Thread(target=begin).start()
time.sleep(2) # Shut down server after 2 seconds
server.shutdown()
The class WSGIRefServer is entirely copied with only 1 line added to the run() method is added. Also add a simple shutdown() method. Unfortunately, this is necessary because of the way bottle creates the run() method.
You can make your thread a daemon by setting the daemon property to True before calling start.
mythread = threading.Thread()
mythread.daemon = True
mythread.start()
A deamon thread will stop whenever the main thread that it is running in is killed or dies. The only problem is that you won't be able to make the thread run any code on exit and if the thread is in the process of doing something, it will be stopped immediately without being able to finish the method it is running.
There's no way in Python to actually explicitly stop a thread. If you want to have more control over being able to stop your server you should look into Python Processes from the multiprocesses module.
Since bottle doesn't provide a mechanism, it requires a hack. This is perhaps the cleanest one if you are using the default WSGI server:
In bottle's code the WSGI server is started with:
srv.serve_forever()
If you have started bottle in its own thread, you can stop it using:
srv.shutdown()
To access the srv variable in your code, you need to edit the bottle source code and make it global. After changing the bottle code, it would look like:
srv = None #make srv global
class WSGIRefServer(ServerAdapter):
def run(self, handler): # pragma: no cover
global srv #make srv global
...
Here's one option: provide custom server (same as default), that records itself:
import bottle
class WSGI(bottle.WSGIRefServer):
instances = []
def run(self, *args, **kw):
self.instances.append(self)
super(WSGI, self).run(*args, **kw)
# some other thread:
bottle.run(host=ip_address, port=12345, server=WSGI)
# control thread:
logging.warn("servers are %s", WSGI.instances)
This is exactly the same method than sepero's and mike's answer, but now much simpler with Bottle version 0.13+:
from bottle import W, run, route
from threading import Thread
import time
#route('/')
def index():
return 'Hello world'
def shutdown():
time.sleep(5)
server.srv.shutdown()
server = WSGIRefServer(port=80)
Thread(target=shutdown).start()
run(server=server)
Also related: https://github.com/bottlepy/bottle/issues/1229 and https://github.com/bottlepy/bottle/issues/1230.
Another example with a route http://localhost/stop to do the shutdown:
from bottle import WSGIRefServer, run, route
from threading import Thread
#route('/')
def index():
return 'Hello world'
#route('/stop')
def stopit():
Thread(target=shutdown).start()
def shutdown():
server.srv.shutdown()
server = WSGIRefServer(port=80)
run(server=server)
PS: it requires at least Bottle 0.13dev.
Console log of Bottle server tells us that the official way of shutting down the server is "Hit Ctrl-C":
Bottle v0.12.19 server starting up (using WSGIRefServer())...
Listening on http://localhost:8080/
Hit Ctrl-C to quit.
Why not simply follow it programmatically?
Hitting "Ctrl-C" is nothing but sending SIGINT to the process, and we can achieve it with just built-in modules:
Get current PID with os.getpid().
Kill the process with os.kill(). Remember passing SIGINT so it will be exactly same as "Hit Ctrl-C".
Wrap the 'kill' in another thread and start it after a few seconds so the client won't get error.
Here is the server code:
from bottle import route, run
import os
import signal
from threading import Thread
import time
#route('/hello')
def return_hello():
return 'Hello'
#route('/stop')
def handle_stop_request():
# Handle "stop server" request from client: start a new thread to stop the server
Thread(target=shutdown_server).start()
return ''
def shutdown_server():
time.sleep(2)
pid = os.getpid() # Get process ID of the current Python script
os.kill(pid, signal.SIGINT)
# Kill the current script process with SIGINT, which does same as "Ctrl-C"
run(host='localhost', port=8080)
Here is the client code:
import requests
def request_service(service_key):
url = f'http://127.0.0.1:8080/{service_key}'
response = requests.get(url)
content = response.content.decode('utf-8')
print(content)
request_service('hello')
request_service('stop')
Note that in function "handle_stop_request" we didn't stop the server immediately but rather started a thread then returned empty string. With this mechanism, when a client requests "http://127.0.0.1:8080/stop", it can get response (the empty string) normally. After that, the server will shutdown. If we otherwise shutdown the server in function "handle_stop_request", the server will close the connection before returning to the client, and hence the client will get "ConnectionError".
Server side output:
Bottle v0.12.19 server starting up (using WSGIRefServer())...
Listening on http://localhost:8080/
Hit Ctrl-C to quit.
127.0.0.1 - - [23/Nov/2021 11:18:08] "GET /hello HTTP/1.1" 200 5
127.0.0.1 - - [23/Nov/2021 11:18:08] "GET /stop HTTP/1.1" 200 0
Client side output:
Hello
The code was tested under Python 3.7 and Bottle 0.12.
I suppose that the bottle webserver runs forever until it terminates. There are no methonds like stop().
But you can make something like this:
from bottle import route, run
import threading, time, os, signal, sys, operator
class MyThread(threading.Thread):
def __init__(self, target, *args):
threading.Thread.__init__(self, target=target, args=args)
self.start()
class Watcher:
def __init__(self):
self.child = os.fork()
if self.child == 0:
return
else:
self.watch()
def watch(self):
try:
os.wait()
except KeyboardInterrupt:
print 'KeyBoardInterrupt'
self.kill()
sys.exit()
def kill(self):
try:
os.kill(self.child, signal.SIGKILL)
except OSError: pass
def background_process():
while 1:
print('background thread running')
time.sleep(1)
#route('/hello/:name')
def index(name='World'):
return '<b>Hello %s!</b>' % name
def main():
Watcher()
MyThread(background_process)
run(host='localhost', port=8080)
if __name__ == "__main__":
main()
Then you can use Watcher.kill() when you need to kill your server.
Here is the code of run() function of the bottle:
try:
app = app or default_app()
if isinstance(app, basestring):
app = load_app(app)
if not callable(app):
raise ValueError("Application is not callable: %r" % app)
for plugin in plugins or []:
app.install(plugin)
if server in server_names:
server = server_names.get(server)
if isinstance(server, basestring):
server = load(server)
if isinstance(server, type):
server = server(host=host, port=port, **kargs)
if not isinstance(server, ServerAdapter):
raise ValueError("Unknown or unsupported server: %r" % server)
server.quiet = server.quiet or quiet
if not server.quiet:
stderr("Bottle server starting up (using %s)...\n" % repr(server))
stderr("Listening on http://%s:%d/\n" % (server.host, server.port))
stderr("Hit Ctrl-C to quit.\n\n")
if reloader:
lockfile = os.environ.get('BOTTLE_LOCKFILE')
bgcheck = FileCheckerThread(lockfile, interval)
with bgcheck:
server.run(app)
if bgcheck.status == 'reload':
sys.exit(3)
else:
server.run(app)
except KeyboardInterrupt:
pass
except (SyntaxError, ImportError):
if not reloader: raise
if not getattr(server, 'quiet', False): print_exc()
sys.exit(3)
finally:
if not getattr(server, 'quiet', False): stderr('Shutdown...\n')
As you can see there are no other way to get off the run loop, except some exceptions.
The server.run function depends on the server you use, but there are no universal quit-method anyway.
This equally kludgy hack has the advantage that is doesn't have you copy-paste any code from bottle.py:
# The global server instance.
server = None
def setup_monkey_patch_for_server_shutdown():
"""Setup globals to steal access to the server reference.
This is required to initiate shutdown, unfortunately.
(Bottle could easily remedy that.)"""
# Save the original function.
from wsgiref.simple_server import make_server
# Create a decorator that will save the server upon start.
def stealing_make_server(*args, **kw):
global server
server = make_server(*args, **kw)
return server
# Patch up wsgiref itself with the decorated function.
import wsgiref.simple_server
wsgiref.simple_server.make_server = stealing_make_server
setup_monkey_patch_for_server_shutdown()
def shutdown():
"""Request for the server to shutdown."""
server.shutdown()
I've found this solution to be the easiest, but it does require that the "psutil" package is installed, to get the current process. It also requires the "signals" module, but that's part of the standard library.
#route('/shutdown')
def shutdown():
current_process = psutil.Process()
current_process.send_signal(signal.CTRL_C_EVENT)
return 'Shutting down the web server'
Hope that's of use to someone!
This question was top in my google search, so i will post my answer:
When the server is started with the Bottle() class, it has a method close() to stop the server. From the source code:
""" Close the application and all installed plugins. """
For example:
class Server:
def __init__(self, host, port):
self._host = host
self._port = port
self._app = Bottle()
def stop(self):
# close ws server
self._app.close()
def foo(self):
# More methods, routes...
Calling stop method will stop de server.