檢測何時按下一個鍵並在按下另一個鍵時中斷[Python]

Question

所以我正在制作一個沒有 PyGame 的游戲，我想添加一個部分，您可以嘗試按給定字母“數字”的正確鍵盤字母，這意味着 A - 1, B - 2, C - 3,等等。我想讓你不能把每把鑰匙都搗碎，所以我加了一個計數器。 但是 - 計數器不起作用。 幫助！

def keyboardpart(l,x,num):
    for i in range(num):
        keypress = False
        c = 0
        dn = random.randint(0,25)
        var = l[dn]
        print(dn+1)
        flag1 = False
        start = time.time()
        while time.time()-start < x:
            if keypress and not keyboard.is_pressed(var):
                if c > 3:
                    break
                c+=1
            elif keyboard.is_pressed(var) and not keypress:
                keypress = True
        print(keypress,c)
        if not keypress:
            print("Sorry, you missed that key.")
            flag1 = True
            break
    if flag1:
        keyboardpart(l,x,num)

Answer 1

檢查是否按下了任何鍵，然后檢查是否按下了目標鍵。 還要等到所有鍵都釋放后再重試。

試試這個代碼：

import random, keyboard, time

l = [chr(65+i) for i in range(26)]  # A-Z

def keyboardpart(l,x,num):
    for i in range(num):
        keyboard.is_pressed(65)  # process OS events
        while len(keyboard._physically_pressed_keys): pass # wait until keys released
        keypress = False
        c = 0
        dn = random.randint(0,25)
        var = l[dn]
        print(dn+1, l[dn])
        flag1 = False
        start = time.time()
        while time.time()-start < x:
            if len(keyboard._physically_pressed_keys):  # any key down
               if keyboard.is_pressed(var):  # check target key
                   keypress = True
                   break
               else:  # wrong key
                   c+=1
                   if c >= 3: break   # allow 3 tries
                   while len(keyboard._physically_pressed_keys): pass # wait until keys released
               print(keypress, c)
        print(keypress,c)
        if not keypress:
            print("Sorry, you missed that key.")
            flag1 = True
            break
             
    if flag1:
        keyboardpart(l,x,num)
        
keyboardpart(l,100,4)