Stopping a Timer in python

Question

I'm making a little gag program that outputs a bunch of random numbers onto my Ubuntu terminal until the number meet a certain condition, at which point access granted is printed.

However, I'd like to stop the Timer loop once that condition is met, and I don't know how.

Here's the code below:

import random
import threading

def message():
    tt = threading.Timer(0.125, message).start()
    num = str(random.randint(137849013724,934234850490))
    print(num)
    if num[3] == "5" and num[6] == "7":
        print("access granted")

message()

If my answer worked for you, please accept it, and consider doing the same for all your previous questions too. — tzaman
– tzaman, Commented Oct 21, 2015 at 6:36

tzaman · Accepted Answer · 2015-10-21 00:39:28Z

3

Timer objects call their function once after a delay, that's it. The reason there's a "loop" is because the function you're telling the timer to call is what's setting up the next timed call to itself. So all you need to do is NOT do that once your condition is met:

def message():
    num = str(random.randint(137849013724,934234850490))
    print(num)
    if num[3] == "5" and num[6] == "7":
        print("access granted")
    else:
        threading.Timer(0.125, message).start()  # <== moved

answered Oct 21, 2015 at 0:39

tzaman

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1 Comment

evandewey Over a year ago

Oops, this comment was intended for the second answer.

TigerhawkT3 · Accepted Answer · 2015-10-21 00:40:04Z

1

You don't need to create a threaded timer for that. Just loop an appropriate quantity of random numbers with a delay:

import time
import random
for i in range(10):
    print(random.randint(1,10), flush=True)
    time.sleep(0.1)

answered Oct 21, 2015 at 0:40

TigerhawkT3

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