python - 检测python中的按键，每次迭代可能需要几秒钟以上？

Question

编辑:以下是使用 的答案keyboard.on_press(回调，抑制=假)在 ubuntu 中运行良好，没有任何问题。
但在 红帽/亚马逊 linux ，它无法工作。

我使用了 thread 中的代码片段

import keyboard  # using module keyboard
while True:  # making a loop
    try:  # used try so that if user pressed other than the given key error will not be shown
        if keyboard.is_pressed('q'):  # if key 'q' is pressed 
            print('You Pressed A Key!')
            break  # finishing the loop
    except:
        break  # if user pressed a key other than the given key the loop will break

但是上面的代码要求每次迭代都在纳秒内执行。在以下情况下失败:

import keyboard  # using module keyboard
import time
while True:  # making a loop
    try:  # used try so that if user pressed other than the given key error will not be shown
        print("sleeping")
        time.sleep(5)
        print("slept")
        if keyboard.is_pressed('q'):  # if key 'q' is pressed 
            print('You Pressed A Key!')
            break  # finishing the loop
    except:
        print("#######")
        break  # if user pressed a key other than the given key the loop will break

Answer 1

您可以在 keyboard 中使用事件处理程序模块以达到预期的效果。

一个这样的处理程序是 keyboard.on_press(callback, suppress=False) :
它为每个 key_down 调用一个回调事件。
您可以引用 keyboard docs

这是您可以尝试的代码:

import keyboard  # using module keyboard
import time

stop = False
def onkeypress(event):
    global stop
    if event.name == 'q':
        stop = True

# ---------> hook event handler
keyboard.on_press(onkeypress)
# --------->

while True:  # making a loop
    try:  # used try so that if user pressed other than the given key error will not be shown
        print("sleeping")
        time.sleep(5)
        print("slept")
        if stop:  # if key 'q' is pressed 
            print('You Pressed A Key!')
            break  # finishing the loop
    except:
        print("#######")
        break  # if user pressed a key other than the given key the loop will break