在python中检测按键，每次迭代可能需要几秒钟以上？

编辑：以下使用键盘的答案。按（callback，suppress=False）在ubuntu中运行良好，没有任何问题。 但在Redhat/Amazon linux中，它无法工作

我已经使用了本文中的代码片段

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

你可以用一根线

import threading

class KeyboardEvent(threading.Thread):
    def run(self):
        if keyboard.is_pressed('q'):  # if key 'q' is pressed 
            print('You Pressed A Key!')
            break  # finishing the loop

keyread = KeyboardEvent()
keyread.start()

---

这将与主线程中的任何内容并行运行，并主要用于侦听按键。

您可以使用

键盘

模块中的事件处理程序来实现所需的结果

一个这样的处理程序是键盘。按下（callback，suppress=False）： 它为每个

按键向下

事件调用回调。 你可以在

以下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

---

编辑：没关系，另一个答案使用了几乎相同的方法

这就是我能想到的，使用相同的“键盘”模块，见下面的代码注释

import keyboard, time
from queue import Queue 

# keyboard keypress callback
def on_keypress(e): 
    keys_queue.put(e.name)

# tell keyboard module to tell us about keypresses via callback
# this callback happens on a separate thread
keys_queue = Queue() 
keyboard.on_press(on_keypress)

try:
    # run the main loop until some key is in the queue
    while keys_queue.empty():  
        print("sleeping")
        time.sleep(5)
        print("slept")
    # check if its the right key
    if keys_queue.get()!='q':
        raise Exception("terminated by bad key")
    # still here? good, this means we've been stoped by the right key
    print("terminated by correct key")
except:
    # well, you can 
    print("#######")
finally:
    # stop receiving the callback at this point
    keyboard.unhook_all()

---

您的程序在睡眠呼叫期间将检测不到任何内容。“你想干什么？”詹姆斯，上面的节目只是一个例子。问题是，如果while循环中的每次迭代都需要几秒钟以上，则不会检测到按键。请检查此示例，您是否尝试使用其他库？就像evemu和python evdev一样，键盘也不一致（至少对我来说）@narenbabr您必须以root用户身份运行此程序，如键盘文档中所述。在运行程序之前，执行sudo su可以解决problem@NarenBabuR是的，那就行了如果我没弄错的话，一旦线程启动，它就会检查按键，然后线程就会退出。你的意思是不是说不是键盘时按了

。而是按了
？
import keyboard, time
from queue import Queue 

# keyboard keypress callback
def on_keypress(e): 
    keys_queue.put(e.name)

# tell keyboard module to tell us about keypresses via callback
# this callback happens on a separate thread
keys_queue = Queue() 
keyboard.on_press(on_keypress)

try:
    # run the main loop until some key is in the queue
    while keys_queue.empty():  
        print("sleeping")
        time.sleep(5)
        print("slept")
    # check if its the right key
    if keys_queue.get()!='q':
        raise Exception("terminated by bad key")
    # still here? good, this means we've been stoped by the right key
    print("terminated by correct key")
except:
    # well, you can 
    print("#######")
finally:
    # stop receiving the callback at this point
    keyboard.unhook_all()