CODEWITHSUNDEEP
pythontimer
evandewey
evandewey

Reputation: 75

Stopping a Timer in python

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()

Upvotes: 0

Views: 1070

Answers (2)

TigerhawkT3
TigerhawkT3

Reputation: 49320

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)

Upvotes: 1

tzaman
tzaman

Reputation: 47770

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

Upvotes: 3

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