Following this example in the ØMQ docs, I'm trying to create a simple receiver. The example uses infinite loop. Everything works just fine. However, on MS Windows, when I hit CTRL+C to raise KeyboardInterrupt, the loop does not break. It seems that recv() method somehow ignores the exception. However, I'd love to exit the process by hiting CTRL+C instead of killing it. Is that possible?
In response to the #Cyclone's request, I suggest the following as a possible solution:
import signal
signal.signal(signal.SIGINT, signal.SIG_DFL);
# any pyzmq-related code, such as `reply = socket.recv()`
A zmq.Poller object seems to help:
def poll_socket(socket, timetick = 100):
poller = zmq.Poller()
poller.register(socket, zmq.POLLIN)
# wait up to 100msec
try:
while True:
obj = dict(poller.poll(timetick))
if socket in obj and obj[socket] == zmq.POLLIN:
yield socket.recv()
except KeyboardInterrupt:
pass
# Escape while loop if there's a keyboard interrupt.
Then you can do things like:
for message in poll_socket(socket):
handle_message(message)
and the for-loop will automatically terminate on Ctrl-C. It looks like the translation from Ctrl-C to a Python KeyboardInterrupt only happens when the interpreter is active and Python has not yielded control to low-level C code; the pyzmq recv() call apparently blocks while in low-level C code, so Python never gets a chance to issue the KeyboardInterrupt. But if you use zmq.Poller then it will stop at a timeout and give the interpreter a chance to issue the KeyboardInterrupt after the timeout is complete.
Don't know if this going to work in Windows, but in Linux I did something like this:
if signal.signal(signal.SIGINT, signal.SIG_DFL):
sys.exit()
Try ctrl+break (as in the key above Page Up, I had to look it up, I don't think I've ever touched that key before)
suggested near the bottom of this thread. I haven't done anything too fancy, but this seems to work well enough in the cases I've tried.
Related
I need to break from time.sleep() using ctrl c.
While 1:
time.sleep(60)
In the above code when the control enters time.sleep function an entire 60 seconds needs to elapsed for python to handled the CTRL C
Is there any elegant way to do it. such that I can interrupt even when the control is in time.sleep function
edit
I was testing it on a legacy implementation which uses python 2.2 on windows 2000 which caused all the trouble . If I had used a higher version of python CTRL C would have interrupted the sleep() . I did a quick hack by calling sleep(1) inside a for loop . which temporarily fixed my issue
The correct answer is to use python stdlib's threading.Event
Sure you can tune down your sleep interval so you sleep for very short periods, but what if you actually want to run your loop once every 60s? Then you need to do more work to determine if it's time to run or just keep sleeping. Furthermore, you're still technically blocking but for only a short period of time. Contrast to threading.Event:
from threading import Event
exit = Event()
def main():
while not exit.is_set():
do_my_thing()
exit.wait(60)
print("All done!")
# perform any cleanup here
def quit(signo, _frame):
print("Interrupted by %d, shutting down" % signo)
exit.set()
if __name__ == '__main__':
import signal
for sig in ('TERM', 'HUP', 'INT'):
signal.signal(getattr(signal, 'SIG'+sig), quit);
main()
When the signal handler calls exit.set(), the main thread's wait() call will immediately be interrupted.
Now, you could use an Event to signal that there's more work to do, etc. But in this case it does double duty as a convenient indicator that we want to quit (e.g. the while not exit.is_set() part.)
You also have the option to put any cleanup code after your while loop.
Not sure what the sense of this code is - but if necessary use a shorter sleep() interval and put a for loop around it:
for i in range(60):
sleep(1)
Catching the KeyboardInterrupt exception using try..except is straight-forward
The KeyboardInterrupt exception is raised when a user hits the interrupt key, Ctrl-C. In python this is translated from a SIGINT signal. That means, you can get handle it however you want using the signal module:
import signal
def handler(signum, frame):
print("do whatever, like call thread.interrupt_main()")
signal.signal(signal.SIGINT, handler)
print("Waiting for SIGINT...")
signal.pause()
That way, you can do whatever you want at the receipt of a keyboard interrupt.
The most elegant solution is certainly threading.Event, though if you only need a quick hack, this code works well :
import time
def main():
print("It’s time !")
if __name__ == "__main__":
print("press ctrl-c to stop")
loop_forever = True
while loop_forever:
main()
try:
time.sleep(60)
except KeyboardInterrupt:
loop_forever = False
I tried your code with python versions 2.5, 2.6, 3 under Linux and all throw "KeyboardInterrupt" exception when hitting CTRL-C.
Maybe some exception handling catches the Interrupt or your problem is like this:
Why is KeyboardInterrupt not working in python?
Based on #Andreas Jung answer
for i in range(360):
try:
sleep(1)
except KeyboardInterrupt:
sys.exit(0)
Figured I'd throw this in.
import time
def sleep(seconds):
for i in range(seconds):
try:
time.sleep(1)
except KeyboardInterrupt:
print("Oh! You have sent a Keyboard Interrupt to me.\nBye, Bye")
break
sleep(60)
In my case, I am handling databases throughout the entire runtime of my program, and so I need to keep my 'cursor' open for the entire program. Is there anyway I can implement a termination protocol, so that when I terminate its execution or an error arises, I am able to run this quick piece of code that simply closes the cursor (I am using python sockets btw).
I would suspect I could do something like this:
if __name__ == "__main__":
Menu()
cursor.close()
However, the only reason that this doesn't work in my case is that Menu is simply starting up threads, and so its execution continues on, returning me back to cursor.close() whilst my program continues to run.
I'm not sure if there is a way to get around this problem.
Yes, you could use the signal library in python to achieve some of this functionality, in particular, capturing program termination as well interrupts to the program like ctrl + c. Example:
# a function to register the signal handlers
# once the program terminates or is halted by an interrupt like ctrl + c it executes the quit_gracefully function
def register_signal_handler():
signal.signal(signal.SIGINT, quit_gracefully)
signal.signal(signal.SIGTERM, quit_gracefully)
return
def quit_gracefully():
# close connections etc.
in case of a different error you could use a try-except block which handles the error and runs the quit_gracefully function in the except.
try:
# some code
except:
quit_gracefully()
EDIT:
this is a good post on signal. How do I capture SIGINT in Python?
You can also use the atexit module: https://docs.python.org/3/library/atexit.html.
Something like this:
import atexit
#atexit.register
def close_cursor():
print("Closing cursor before exiting.")
cursor.close()
I currently have code that basically runs an infinite while loop to collect data from users. Constantly updating dictionaries/lists based on the contents of a text file. For reference:
while (True):
IDs2=UpdatePoints(value,IDs2)
time.sleep(10)
Basically, my problem is that I do not know when I want this to end, but after this while loop runs I want to use the information collected, not lose it by crashing my program. Is there a simple, elegant way to simply exit out of the while loop whenever I want? Something like pressing a certain key on my keyboard would be awesome.
You can try wrapping that code in a try/except block, because keyboard interrupts are just exceptions:
try:
while True:
IDs2=UpdatePoints(value,IDs2)
time.sleep(10)
except KeyboardInterrupt:
print('interrupted!')
Then you can exit the loop with CTRL-C.
You could use exceptions. But you only should use exceptions for stuff that isn't supposed to happen. So not for this.
That is why I recommand signals:
import sys, signal
def signal_handler(signal, frame):
print("\nprogram exiting gracefully")
sys.exit(0)
signal.signal(signal.SIGINT, signal_handler)
you should put this on the beginning of your program and when you press ctrl+c wherever in your program it will shut down gracefully
Code explanation:
You import sys and signals.
Then you make a function that executes on exit. sys.exit(0) stops the programming with exit code 0 (the code that says, everything went good).
When the program get the SIGINT either by ctrl-c or by a kill command in the terminal you program will shutdown gracefully.
I think the easiest solution would be to catch the KeyboardInterrupt when the interrupt key is pressed, and use that to determine when to stop the loop.
except KeyboardInterrupt:
break
The disadvantage of looking for this exception is that it may prevent the user from terminating the program while the loop is still running.
I use python to track stock prices and submit automated buy/sell commands on my portfolio. Long story short, I wanted my tracking program to ping the data server for info, and place trades off of the information gathered, but I also wanted to save the stock data for future reference, on top of being able to start/stop the program whenever I wanted.
What ended up working for me was the following:
trigger = True
while trigger == True:
try:
(tracking program and purchasing program conditions here)
except:
trigger = False
print('shutdown initialized')
df = pd.DataFrame...
save all the datas
print('shutdown complete')
etc.
From here, while the program is in the forever loop spamming away requests for data from my broker's API, using the CTRL-C keyboard interrupt function toggles the exception to the try loop, which nullifies the while loop, allowing the script to finalize the data saving protocol without bringing the entire script to an abrupt halt.
Hope this helps!
Resultant
I would suggest using the try, except syntax within a loop if you are running on an IPYNB file in Google Colab or Jupyter, like:
while True:
try:
IDs2=UpdatePoints(value,IDs2)
time.sleep(10)
except KeyboardInterrupt:
break
except:
continue
the last except is for any other error if occurs the loop will resume
You can catch the KeyboardInterrupt error in Python:
try:
while 1>0:
IDs2=UpdatePoints(value,IDs2)
time.sleep(10)
except KeyboardInterrupt:
print('While loop ended!')
Also, instead of saying:
while True:
It looks more professional to use:
while 1>0:
To read more about Python Error handling (try, except, etc.):
https://www.w3schools.com/python/python_try_except.asp
or:
https://www.w3schools.com/python/gloss_python_try_finally.asp
I've read about this technique to timeout blocking IO operations, the problem is that it doesn't seem to work. for example:
import thread, threading
def read_timeout(prompt, timeout=10.0):
timer = threading.Timer(timeout, thread.interrupt_main)
s = ''
timer.start()
try:
s = raw_input(prompt)
except KeyboardInterrupt:
print 'operation timed out.'
timer.cancel()
return s
s = read_timeout('enter input: ')
if s:
print 'you entered: %s' % s
this won't interrupt the main thread until raw_input() returns.
Any help is appreciated.
Update:
Using os.kill(os.getpid(), signal.SIGINT) instead of thread.interrupt_main() seems to work (at least on Linux, which doesn't give me the portability I initially wanted). However, I'm still wondering why the code above doesn't work.
On Unix machines, there is a way to do what you are trying to do. Look at this post:
raw_input and timeout
Just remove the comma at the end of line 5, or the prompt won't be displayed until the program is terminated.
On the same page there's also solution for Windows OS, but I haven't tested it to know if it's working.
According to the latest doc of the API: https://docs.python.org/3/library/_thread.html#thread.interrupt_main
This does not emit the corresponding signal but schedules a call to the associated handler (if it exists).
And the call is performed after the current blocking call in the main thread.
On the contrary, the os.kill solution mentioned in the update actually emits a signal to the main thread.
Sorry that it's been quite a while since the question was raised, but hopefully it still helps.
My code looks like this:
... # class Site(Resource)
def render_POST(self,request)
otherclass.doAssync(request.args)
print '1'
return "done" #that returns the HTTP response, always the same.
...
def doAssync(self,msg):
d = defer.Deferred()
reactor.callLater(0,self.doStuff,d,msg)
d.addCallback(self.sucess)
def doStuff(self,d,msg):
# do some stuff
time.sleep(2) #just for example
d.callback('ok')
def sucess(msg):
print msg
The output:
1
ok
So far, so good, but, the HTTP response (return 'done'), only happens after the delay (time.sleep(2)).
I can tell this, because the browser keeps 'loading' for 2 seconds.
What am I doing wrong?
What you are doing wrong is running a blocking call (time.sleep(2)), while Twisted expects you to only perform non-blocking operations. Things that don't wait. Because you have that time.sleep(2) in there, Twisted can't do anything else while that function is sleeping. So it can't send any data to the browser, either.
In the case of time.sleep(2), you would replace that with another reactor.callLater call. Assuming you actually meant for the time.sleep(2) call to be some other blocking operation, how to fix it depends on the operation. If you can do the operation in a non-blocking way, do that. For many such operations (like database interaction) Twisted already comes with non-blocking alternatives. If the thing you're doing has no non-blocking interface and Twisted doesn't have an alternative to it, you may have to run the code in a separate thread (using for example twisted.internet.threads.deferToThread), although that requires your code is actually thread-safe.