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我想通过 AutoKey 获得按下的号码。我的脚本可以工作，但速度很慢，而且看起来不太好。

你能知道一个更快的方法吗？我需要在它被识别前几秒钟按住一个键。

import os, time, subprocess 
def popupNotify(text):
    subprocess.Popen(['notify-send', text])  # will be showed right top
pressed_key = 999999999999
for x in range(0, 150):
    retCode1 = keyboard.wait_for_keypress('<np_end>',modifiers=[],timeOut=0.01) # <== works
    retCode2 = keyboard.wait_for_keypress('<np_down>',modifiers=[],timeOut=0.01) # <== works
    retCode3 = keyboard.wait_for_keypress('<np_page_down>',modifiers=[],timeOut=0.01) # <== works
    retCode4 = keyboard.wait_for_keypress('<np_left>',modifiers=[],timeOut=0.001) # <== works
    #retCode5 = keyboard.wait_for_keypress('5',modifiers=[],timeOut=0.001) # <== works
    #retCode5 = keyboard.wait_for_keypress('<code84>',modifiers=[],timeOut=0.001) # <== not works, no error
    if retCode1:
        pressed_key = 1
    if retCode2:
        pressed_key = 2
    if retCode3:
        pressed_key = 3
    if retCode4:
        pressed_key = 4
    if pressed_key != 999999999999:
        break

popupNotify(str(pressed_key))
popupNotify("END END END END ")

我在这里读到：

系统

AutoKey (Qt) 0.95.10
Python 3.8.5
Operating System: Kubuntu 20xx
KDE Plasma Version
4

1 回答 1

1

如果您想使用自动键从用户那里获取输入，我认为最好的方法是打开一个对话框：

import subprocess
a = dialog.input_dialog(title='Enter a value', message='Enter a value', default='')
subprocess.Popen(['notify-send', a.data])  # will be showed right top
于 2021-05-06T15:07:35.960 回答