python - 缩短用户输入的 sleep 时间

Question

我试图找到一种方法来缩短 time.sleep(600)如果用户输入一个键，而不诉诸于一些丑陋的黑客，如:

key_pressed = False
for i in range(600):
    key_pressed = key_was_pressed()
    if not key_pressed:
        time.sleep(1)
    else:
        break

Answer 1

这是使用 signal.alarm 的实现的跨平台改编。中断(在 Windows 上不可用的想法)。此代码应该适用于 Linux、macOS 和 Windows。第三方辅助库 readchar 可以用pip install readchar安装.

import os
import signal
import sys

from threading import Timer
from readchar import readkey

def wait_for(key="x", timeout=600):
    pid = os.getpid()
    sig = signal.CTRL_C_EVENT if os.name == "nt" else signal.SIGINT
    timer = Timer(timeout, lambda: os.kill(pid, sig))
    print(f"waiting {timeout}s for user to press {key!r} ...")
    timer.start()  # spawn a worker thread to interrupt us later
    while True:
        k = readkey()
        print(f"received {k!r}")
        if k == key:
            timer.cancel()  # cancel the timer
            print("breaking")
            break

def main():
    import sys
    try:
        wait_for(key=sys.argv[1], timeout=int(sys.argv[2]))
    except KeyboardInterrupt as err:
        print("user took too long")

if __name__ == "__main__":
    main()