问题:在python中检测按键?
我正在用python开发一个秒表类型的程序,我想知道如何检测是否按下了一个键(例如p表示暂停,s表示停止),而我不希望它像raw_input这样等待用户输入,然后继续执行。有人知道如何在while循环中执行此操作吗?
另外,我想做这个跨平台的,但是如果那不可能,那么我的主要开发目标是linux
回答 0
Python有一个具有许多功能的键盘模块。安装它,也许使用以下命令:
pip3 install keyboard
然后在如下代码中使用它:
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
回答 1
from pynput.keyboard import Key, Listener
def on_press(key):
print('{0} pressed'.format(
key))
def on_release(key):
print('{0} release'.format(
key))
if key == Key.esc:
# Stop listener
return False
# Collect events until released
with Listener(
on_press=on_press,
on_release=on_release) as listener:
listener.join()
上面的功能将打印您所按的任何键,并在您释放“ esc”键时启动操作。键盘文档在这里用于更多变化的用法。
马库斯·冯·布罗迪(Markus von Broady)强调了一个潜在的问题,即:这个答案并不需要您在当前窗口中激活此脚本,Windows的解决方案是:
from win32gui import GetWindowText, GetForegroundWindow
current_window = (GetWindowText(GetForegroundWindow()))
desired_window_name = "Stopwatch" #Whatever the name of your window should be
#Infinite loops are dangerous.
while True: #Don't rely on this line of code too much and make sure to adapt this to your project.
if current_window == desired_window_name:
with Listener(
on_press=on_press,
on_release=on_release) as listener:
listener.join()
回答 2
正如OP关于raw_input的提及-这意味着他想要cli解决方案。Linux:curses是您想要的(Windows PDCurses)。Curses是cli软件的图形API,您不仅可以检测关键事件,还可以实现更多目标。
该代码将检测按键,直到按下新行为止。
import curses
import os
def main(win):
win.nodelay(True)
key=""
win.clear()
win.addstr("Detected key:")
while 1:
try:
key = win.getkey()
win.clear()
win.addstr("Detected key:")
win.addstr(str(key))
if key == os.linesep:
break
except Exception as e:
# No input
pass
curses.wrapper(main)
回答 3
使用keyboard
模块可以做更多的事情。
以下是一些方法:
方法1:
使用功能read_key()
:
import keyboard
while True:
if keyboard.read_key() == "p":
print("You pressed p")
break
p按下该键将打破循环。
方法2:
使用功能wait
:
import keyboard
keyboard.wait("p")
print("You pressed p")
它将等待您按下p并继续按下代码。
方法3:
使用功能on_press_key
:
import keyboard
keyboard.on_press_key("p", lambda _:print("You pressed p"))
它需要一个回调函数。我_
之所以使用,是因为键盘功能将键盘事件返回到该功能。
一旦执行,当按下键时它将运行该功能。您可以通过运行以下行来停止所有挂钩:
keyboard.unhook_all()
方法4:
user8167727已经回答了这种方法,但是我不同意他们编写的代码。它将使用该函数,is_pressed
但以另一种方式:
import keyboard
while True:
if keyboard.is_pressed("p"):
print("You pressed p")
break
p按下将打破循环。
笔记:
keyboard
将从整个操作系统读取按键。keyboard
在Linux上需要root
回答 4
对于Windows,您可以这样使用msvcrt
:
import msvcrt
while True:
if msvcrt.kbhit():
key = msvcrt.getch()
print(key) # just to show the result
回答 5
使用此代码查找按下了哪个键
from pynput import keyboard
def on_press(key):
try:
print('alphanumeric key {0} pressed'.format(
key.char))
except AttributeError:
print('special key {0} pressed'.format(
key))
def on_release(key):
print('{0} released'.format(
key))
if key == keyboard.Key.esc:
# Stop listener
return False
# Collect events until released
with keyboard.Listener(
on_press=on_press,
on_release=on_release) as listener:
listener.join()
回答 6
使用PyGame拥有一个窗口,然后您可以获取关键事件。
对于这封信p
:
import pygame, sys
import pygame.locals
pygame.init()
BLACK = (0,0,0)
WIDTH = 1280
HEIGHT = 1024
windowSurface = pygame.display.set_mode((WIDTH, HEIGHT), 0, 32)
windowSurface.fill(BLACK)
while True:
for event in pygame.event.get():
if event.key == pygame.K_p: # replace the 'p' to whatever key you wanted to be pressed
pass #Do what you want to here
if event.type == pygame.locals.QUIT:
pygame.quit()
sys.exit()
回答 7
所以我根据这篇文章(使用msvcr库和Python 3.7)制作了这款游戏。
以下是游戏的“主要功能”,即检测所按下的按键:
# Requiered libraries - - - -
import msvcrt
# - - - - - - - - - - - - - -
def _secret_key(self):
# Get the key pressed by the user and check if he/she wins.
bk = chr(10) + "-"*25 + chr(10)
while True:
print(bk + "Press any key(s)" + bk)
#asks the user to type any key(s)
kp = str(msvcrt.getch()).replace("b'", "").replace("'", "")
# Store key's value.
if r'\xe0' in kp:
kp += str(msvcrt.getch()).replace("b'", "").replace("'", "")
# Refactor the variable in case of multi press.
if kp == r'\xe0\x8a':
# If user pressed the secret key, the game ends.
# \x8a is CTRL+F12, that's the secret key.
print(bk + "CONGRATULATIONS YOU PRESSED THE SECRET KEYS!\a" + bk)
print("Press any key to exit the game")
msvcrt.getch()
break
else:
print(" You pressed:'", kp + "', that's not the secret key(s)\n")
if self.select_continue() == "n":
if self.secondary_options():
self._main_menu()
break
如果您想要该porgram的完整源代码,则可以在这里查看或下载它:
(注意:秘密按键为:Ctrl+ F12)
我希望您可以作为一个例子,并为那些来此信息的人提供帮助。
回答 8
我建议您使用PyGame并添加一个事件句柄。
回答 9
key = cv2.waitKey(1)
这是来自openCV软件包。它无需等待即可检测到按键。