如何使用curses按下最后一个箭头键？

发布于 2024-09-18 01:17:44 字数 572 浏览 11 评论 0原文

我正在编写一个使用诅咒的 Python 蛇游戏，但在控制蛇时遇到一些问题，我当前的代码用于控制蛇的内容放置在主循环内，如下所示：

while True:
    char = screen.getch()
    if char == 113: exit()  # q
    elif char == curses.KEY_RIGHT: snake.update(RIGHT)
    elif char == curses.KEY_LEFT: snake.update(LEFT)
    elif char == curses.KEY_UP: snake.update(UP)
    elif char == curses.KEY_DOWN: snake.update(DOWN)
    else snake.update()
    time.sleep(0.1)

但是，代码似乎将按下的键视为 que（因此蛇在用完箭头按下时会停止），而我实际上希望它检索最后按下的箭头键。

如何检索最后按下的箭头键？

原文

I'm writing a Python snake game using curses, but am having some trouble controlling the snake, my current code for controlling the snake is placed inside the main loop and looks like this:

while True:
    char = screen.getch()
    if char == 113: exit()  # q
    elif char == curses.KEY_RIGHT: snake.update(RIGHT)
    elif char == curses.KEY_LEFT: snake.update(LEFT)
    elif char == curses.KEY_UP: snake.update(UP)
    elif char == curses.KEY_DOWN: snake.update(DOWN)
    else snake.update()
    time.sleep(0.1)

However the code seems to treat the keys pressed as a que (so the snake will stop when it runs out of arrow-presses), whereas I actually want it to retrieve the last arrow key that was pressed.

How can I retrieve the last arrow key that was pressed?

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人生戏 2024-09-25 01:17:44

设置 screen.nodelay(1)：

screen.nodelay(1)
while True:
    char = screen.getch()
    if char == 113: break  # q
    elif char == curses.KEY_RIGHT: snake.update(RIGHT)
    elif char == curses.KEY_LEFT: snake.update(LEFT)
    elif char == curses.KEY_UP: snake.update(UP)
    elif char == curses.KEY_DOWN: snake.update(DOWN)
    else: snake.update()
    time.sleep(0.1)

Set screen.nodelay(1):

screen.nodelay(1)
while True:
    char = screen.getch()
    if char == 113: break  # q
    elif char == curses.KEY_RIGHT: snake.update(RIGHT)
    elif char == curses.KEY_LEFT: snake.update(LEFT)
    elif char == curses.KEY_UP: snake.update(UP)
    elif char == curses.KEY_DOWN: snake.update(DOWN)
    else: snake.update()
    time.sleep(0.1)

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