我要通过AutoKey按下电话号码。我的脚本工作,但真的很慢,看起来不太好。
你能知道更快的方法吗?我需要在钥匙被识别前几秒钟按住它。
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发布于 2021-05-06 15:07:35
如果您想使用自动键从用户获得输入,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 tophttps://stackoverflow.com/questions/65589400
复制相似问题