Question

点击按钮序列，例如 shift + a 并打开amazon.com。单击另一个按钮序列，例如 shift + e 并打开asyncio.create_task。该代码适用于ebay.com，但是当我单击热键 shift + e 时，它没有打开ebay.com。当您打印要单击的键时，您会意识到该程序正在按 Shift ， e 和 a 。我不知道为什么程序会打印我不单击的 a ？

amazon.com

Answer 1

首先，最好在组合后清除current，而不要在密钥释放后删除项目，这样会更好，更简单。第二个all(k in current for k in COMBO)仅检查COMBO中的所有元素是否都位于current中，而不检查方向。例如，如果您按下 a + Shift ，它将变成True，这是错误的。一种解决方案是使用list代替set，并检查COMBO中current是否存在（作为切片）。我用this example检查切片列表的存在。

import time
from pynput import keyboard
COMBINATIONS = [
    [keyboard.Key.shift, keyboard.KeyCode(char='a')],
    [keyboard.Key.shift, keyboard.KeyCode(char='A')],
    [keyboard.Key.shift, keyboard.KeyCode(char='e')],
    [keyboard.Key.shift, keyboard.KeyCode(char='E')]
]
current = []


def execute(url):
    keyboard_ctrl = keyboard.Controller()
    keyboard_ctrl.press(keyboard.Key.ctrl_l)
    keyboard_ctrl.press('l')
    keyboard_ctrl.release(keyboard.Key.ctrl_l)
    keyboard_ctrl.release('l')
    time.sleep(0.2)

    for i in url:
        keyboard_ctrl.press(i)
        keyboard_ctrl.release(i)
    # keyboard_ctrl.type(url)
    time.sleep(0.2)
    keyboard_ctrl.press(keyboard.Key.enter)
    keyboard_ctrl.release(keyboard.Key.enter)


def contains_sublist(lst, sublst):
    n = len(sublst)
    return any((sublst == lst[i:i+n]) for i in range(len(lst)-n+1))


def on_press(key):
    global current
    if any([key in COMBO for COMBO in COMBINATIONS]):
        current.append(key)
        print(key)
        if any((contains_sublist(current, COMBO)) for COMBO in COMBINATIONS):
            print("Combination pressed!")
            if current[-1] == keyboard.KeyCode(char='a'):
                execute('https://www.amazon.com/')
            if current[-1] == keyboard.KeyCode(char='e'):
                execute('https://www.ebay.com/')
            current = []


with keyboard.Listener(on_press=on_press) as listener:
    listener.join()

如何单击热键并在python中打开特定的URL？

1 个答案: