Python嵌套条件？双重条件后跟另一个条件

Question

我正在尝试学习 python，但在这个问题上苦苦挣扎。

想象一下这种情况：

1- 用户按下按钮 1 和按钮 2 = 程序需要“做一些”

2 - 用户按下按钮 1 和按钮 2，然后按下按钮 3 = 我可以“做一些 2”

3 - 用户按下按钮 2 和按钮 1，然后按下按钮 3 = 我可以“做一些 2”

但是

4 - 如果用户在 button1 和 button2 之前按下 button3 = “do some2”对我来说不行，我只需要“do some”

所以首先按下哪个按钮的顺序对我的目的很重要

def button1_pressed():  # Returns true if the left mouse button is pressed
    button1_state = win32api.GetKeyState(0x01)
    return button1_state < 0

def button2_pressed():  # Returns true if the right mouse button is pressed
    button2_state = win32api.GetKeyState(0x02)
    return  button2_state < 0

def button3_pressed():  # Returns true if the M button is pressed
    button3_state = win32api.GetKeyState(0x4D)
    return button3_state < 0

while True:
    while button1_pressed() and button2_pressed(): # No matter which one is pressed first to me
        print("do some")
        while button3_pressed(): # Only if it is pressed after button1 and button2 are pressed
            print("do some2")
            if button2_pressed() == 0:
                break # this only breaks the inner while loop
        break # added an extra break else it would just be stuck in an endless loop

如果有人给我一些帮助，我将不胜感激。

Answer 1

您在这里想要的是所谓的“状态机”。您描述的情况称为“状态”，并且在按钮按下停止后它们会持续存在（例如，您可以处于“用户按下按钮 2”的状态，这与“用户正在按下按钮 2”不同）。

pressed = [True, False, False, False]
while True:
  if button1_pressed() and not pressed[1]:
      pressed[1] = True
      if pressed[2]:
          print("do some")
          continue
  if button2_pressed() and not pressed[2]:
      pressed[2] = True
      if pressed[1]:
          print("do some")
          continue
  if button3_pressed() and not pressed[3]:
      pressed[3] = True
      if pressed[1] and pressed[2]:
          print("do some 2")
  if all(pressed):
      # All the buttons have been pressed (in some order).  Done!
      break

您可能会在数据结构中对其进行编码，如下所示：

buttons = [lambda: False, button1_pressed, button2_pressed, button3_pressed]
pressed = [False, False, False, False]
actions = {
    # For each button press, what action happens 
    # based on button(s) previously pressed.
    1: ([2], "do some")
    2: ([1], "do some")
    3: ([1, 2], "do some 2")
}
while True:
    for num in (1, 2, 3):
        if not buttons[num]() or pressed[num]:
            continue
        pressed[num] = True
        prev, action = actions[num]
        if all(pressed[p] for p in prev):
            print(action)
    if any(
        not pressed_now() and was_pressed
        for pressed_now, was_pressed in zip(buttons, pressed)
    ):
        # Break condition: previously pressed button was released
        break