如何在pygame中上下移动一个方框[矩形为方形]?

时间:2017-08-13 10:27:51

标签: python python-3.x pygame

我正试图在pygame中上下移动一个方框 我可以使用键a向左移动框,并使用键d向右移动 我怎样才能上下移动?
我的代码:

import sys
import pygame as pg


def main():
    screen = pg.display.set_mode((640, 480))
    clock = pg.time.Clock()
    rect = pg.Rect(300, 220, 20, 20)
    velocity = (0, 0)
    done = False

    while not done:
        for event in pg.event.get():
            if event.type == pg.QUIT:
                done = True

        keys = pg.key.get_pressed()
        if keys[pg.K_a]:  #to move left
            rect.x -= 4
        if keys[pg.K_d]: #to move right
            rect.x += 4


        screen.fill((40, 40, 40))
        pg.draw.rect(screen, (150, 200, 20), rect)

        pg.display.flip()
        clock.tick(30)


if __name__ == '__main__':
    pg.init()
    main()
    pg.quit()
    sys.exit()

2 个答案:

答案 0 :(得分:1)

您可以使用与水平移动相同的逻辑,方法是更改​​y坐标:

        y-4
         |
x-4 <-- [_] --> x+4
         |
        y+4


if keys[pg.K_w]: # to move up
    rect.y -= 4
if keys[pg.K_s]: # to move down
    rect.y += 4

答案 1 :(得分:-1)

我认为按下方向键会更有意义移动方框。

(内部未完成:循环)

if event.type == pg.KEYDOWN:
    if event.key == pg.K_LEFT:
        rect.x -= 4
    if event.key == pg.K_RIGHT:
        rect.x += 4
    if event.key == pg.K_UP:
        rect.y -= 4
    if event.key == pg.K_DOWN:
        rect.y += 4