問題描述

我正在編寫一個腳本，它會截取屏幕截圖并解碼以圖像名稱命名的特定按鍵，如下所示.我的問題是，當我按下左鍵盤箭頭時，也按下了數字 4.我在谷歌或鍵盤庫的文檔中找不到任何東西.我正在使用 Windows 和 Python 3.6.5

I was writing a script, which takes a screenshot and decodes specific key presses in the name of the image as seen below. My problem is that when I press the left keyboard arrow, also the number 4 is pressed. I can't find anything on google or in the documentation of the keyboard library. I am using Windows and Python 3.6.5

(75,)
left arrow pressed
(5, 75)
4 pressed

向下箭頭也會發生同樣的事情，但數字 3 是這樣的.

The same thing happens with the down arrow, but with the number 3.

(80,)
down arrow pressed
(3, 80)
2 pressed

代碼:

from PIL import ImageGrab
import keyboard  # using module keyboard
import time

keys = [
    "down arrow",
    "up arrow",
    "left arrow",
    "right arrow",
    "w",
    "s",
    "a",
    "d",
    "1",
    "2",
    "3",
    "4",
    "q",
    

e",f"]

if __name__ == "__main__":
    while True:
        code = []
        try:
            for key in keys:
                if keyboard.is_pressed(key):
                    print(keyboard.key_to_scan_codes(key))
                    print(f"{key} pressed")
                    code.append(1)
                else:
                    code.append(0)
                    
            if keyboard.is_pressed('esc'):
                print(key + " pressed")
                break
                
            c = "".join(map(str, code))
            snapshot = ImageGrab.grab()
            save_path = str(int(time.time()*1000)) + "-" + c + ".jpg"
            snapshot.save("tmp\" + save_path)

        except:
            break

推薦答案

keyboard 模塊對于此類實例有簡單的解決方案，它們使用 event-triggered 激活而不是polling 在您的嘗試中使用.

The keyboard module has simple solutions for instances like these, they use event-triggered activation rather than polling as is used in your attempt.

示例代碼:

import keyboard

def handleLeftKey(e):
    if keyboard.is_pressed("4"):
        print("left arrow was pressed w/ key 4")
        # work your magic

keyboard.on_press_key("left", handleLeftKey)
# self-explanitory: when the left key is pressed down then do something

keyboard.on_release_key("left", handleLeftKey02)
# also self-explanitory: when the left key is released then do something

# don't use both ...on_release & ...on_press or it will be
# triggered twice per key-use (1 up, 1 down)

替換下面的代碼并根據您的需要進行更改.

Replace the code below and change it to suit your needs.

if __name__ == "__main__":
    while True:
        code = []
        try:
            for key in keys:
                if keyboard.is_pressed(key):
                    print(keyboard.key_to_scan_codes(key))
                    print(f"{key} pressed")
                    code.append(1)
                else:
                    code.append(0)

另一種更動態的方法如下所示:

Another, more dynamic approach would look like:

import keyboard

keys = [
    "down",
    "up",
    "left",
    "right",
    "w",
    "s",
    "a",
    "d",
    "1",
    "2",
    "3",
    "4",
    "q",
    "e",
    "f"
]

def kbdCallback(e):
    found = False
    for key in keys:
        if key == keyboard.normalize_name(e.name):
            print(f"{key} was pressed")
            found = True
            # work your magic

    if found == True:
        if e.name == "left":
            if keyboard.is_pressed("4"):
                print("4 & left arrow were pressed together!")
                # work your magic

keyboard.on_press(kbdCallback)
# same as keyboard.on_press_key, but it does this for EVERY key

我注意到的另一個問題是您使用 "left arrow" 而實際上它被識別為 "left" (至少在我的系統上，它可能會有所不同在你的，但我假設你希望它在所有系統上工作，所以使用 "left" 會更安全)

Another issue I noticed was that you were using "left arrow" when really it was recognized as "left" (at least on my system, it may be different on yours, but I assume you want it to work on all systems so it'd be safer using "left" instead)

您可以使用的最后一種方法是非常靜態類型的并且沒有動態功能，但可以在 "4+left" 或 "left+4"

The last method you could use is very statically typed and has no dynamic capabilities, but would work in the case of "4+left" or "left+4"

import keyboard

def left4trigger:
    print("the keys were pressed")

keyboard.add_hotkey("4+left", left4trigger)
# works as 4+left or left+4 (all of the examples do)

你看起來很聰明，可以從那里弄清楚其余的事情.

You seem smart enough to figure out the rest from there.

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