在Pygame中搭建平台

Question

我试图让我的平台游戏，不知道如何让玩家不通过平台。我对Python和Pygame非常陌生，因此如果能得到一些帮助，我将不胜感激。到目前为止，我已经添加了动画、背景和跳转机制。我主要遵循了一个指南，但也自己添加了一些东西。

import pygame

pygame.init()

win = pygame.display.set_mode((500, 480))
pygame.display.set_caption("First Game")

walkRight = [pygame.image.load('R1.png'), pygame.image.load('R2.png'), pygame.image.load('R3.png'),
             pygame.image.load('R4.png'), pygame.image.load('R5.png'), pygame.image.load('R6.png'),
             pygame.image.load('R7.png'), pygame.image.load('R8.png'), pygame.image.load('R9.png')]
walkLeft = [pygame.image.load('L1.png'), pygame.image.load('L2.png'), pygame.image.load('L3.png'),
            pygame.image.load('L4.png'), pygame.image.load('L5.png'), pygame.image.load('L6.png'),
            pygame.image.load('L7.png'), pygame.image.load('L8.png'), pygame.image.load('L9.png')]
bg = pygame.image.load('bg.jpg')
char = pygame.image.load('standing.png')
platform = pygame.image.load('platform.png')

x = 50
y = 400
width = 40
height = 60
vel = 5
platform_x = 250
platform_y = 320

clock = pygame.time.Clock()

isJump = False
jumpCount = 10

left = False
right = False
walkCount = 0
platform_pos = 320





def redrawGameWindow():
    global walkCount

    win.blit(bg, (0, 0))
    if walkCount + 1 >= 27:
        walkCount = 0

    if left:
        win.blit(walkLeft[walkCount // 3], (x, y))
        walkCount += 1
    elif right:
        win.blit(walkRight[walkCount // 3], (x, y))
        walkCount += 1
    else:
        win.blit(char, (x, y))
        walkCount = 0
    win.blit(platform, (platform_x,platform_y))

    pygame.display.update()


run = True

while run:
    clock.tick(27)

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            run = False

    keys = pygame.key.get_pressed()

    if keys[pygame.K_LEFT] and x > vel:
        x -= vel
        left = True
        right = False

    elif keys[pygame.K_RIGHT] and x < 500 - vel - width:
        x += vel
        left = False
        right = True

    else:
        left = False
        right = False
        walkCount = 0

    if not isJump:
        if keys[pygame.K_SPACE]:
            isJump = True
            left = False
            right = False
            walkCount = 0
    else:
        if jumpCount >= -10:
            y -= (jumpCount * abs(jumpCount)) * 0.5
            jumpCount -= 1
        else:
            jumpCount = 10
            isJump = False






    redrawGameWindow()

pygame.quit()

Answer 1

我处理这个问题的方式是将平台创建为pygame矩形，self.platform = pygame.Rect(x, y, w, w)。然后，我在播放器周围创建了一个hitbox，例如一个类似于前面的矩形：self.player_hitbox = pygame.Rect(x, y, w, w)。

然后，您可以使用代码if self.player_hitbox.colliderect(self.platform)检查与2的冲突，并对该信息执行您想要的操作。