Question

我一直在编写一个简单的平台游戏，用户控制一个正方形并且必须到达窗口的另一侧，而不会与其他正方形碰撞。我希望其他4个方块上下反弹，所以我写了这段代码：

import pygame
import os
import sys

os.environ['SDL_VIDEO_CENTERED'] = "1"
pygame.init()

#Variables:
width = 600
height = width / 16 * 9

running = True

#Colors
PINK = (255, 79, 161)
BLACK = (0, 0, 0)
BLUE = (0, 0, 255)

clock = pygame.time.Clock()

#MainRectProperties
mainRectX = 0
mainRectY = height / 2 - 20
mainRectSpeed = 250

#RectOneProperties:
rectOneX = 150
rectOneY = 0

#RectTwoPropeties:
rectTwoX = 250
rectTwoY = height - 20

#RectThreeProperties:
rectThreeX = 350
rectThreeY = 0

#RectFourProperties:
rectFourX = 450
rectFourY = height - 20

#Window:
window = pygame.display.set_mode((width, height))
windowText = pygame.display.set_caption("Pixel Animation")

#Rectangles:
mainRect = pygame.draw.rect(window, PINK, (mainRectX, mainRectY, 20, 20), 0)
obstacleRect1 = pygame.draw.rect(window, BLUE, (rectOneX, rectOneY, 20, 20), 0)
obstacleRect2 = pygame.draw.rect(window, BLUE, (rectTwoX, rectTwoY, 20, 20,), 0)
obstacleRect3 = pygame.draw.rect(window, BLUE, (rectThreeX, rectThreeY, 20, 20), 0)
obstacleRect4 = pygame.draw.rect(window, BLUE, (rectFourX, rectFourY, 20, 20), 0)
pygame.display.flip()

#UpdateMainRectFunction
def updateMainRect(x, y):
    window.fill(BLACK)
    mainRect = pygame.draw.rect(window, PINK, (mainRectX, mainRectY, 20, 20), 0)
    pygame.display.flip()
    clock.tick(250)

#GameLoop
while running:
    goingDown = True
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            sys.exit()

    if rectOneY < height - 21 and goingDown:
        print "b"
        pygame.draw.rect(window, BLACK, (rectOneX, rectOneY, 20, 20), 0)
        rectOneY += 1
        obstacleRect1 = pygame.draw.rect(window, BLUE, (rectOneX, rectOneY, 20, 20), 0)
        pygame.display.flip()
        clock.tick(100)
        if rectOneY == height - 21:
            goingDown = False
        else:
            goingDown = True

    if not goingDown and rectOneY != 0:
        print "a"
        pygame.draw.rect(window, BLACK, (rectOneX, rectOneY, 20, 20), 0)
        rectOneY -= 1
        obstacleRect1 = pygame.draw.rect(window, BLUE, (rectOneX, rectOneY, 20, 20), 0)
        pygame.display.flip()
        clock.tick(100)
        goingDown = False
        print rectOneY < height - 21 and goingDown     


    keys = pygame.key.get_pressed()
    #MovingRectCommands
    if keys[pygame.K_UP]:
        mainRectY -= 1
        updateMainRect(mainRectX, mainRectY)

    if keys[pygame.K_DOWN]:
        mainRectY += 1
        updateMainRect(mainRectX, mainRectY)

    if keys[pygame.K_LEFT]:
        mainRectX -= 1
        updateMainRect(mainRectX, mainRectY)

    if keys[pygame.K_RIGHT]:
        mainRectX += 1
        updateMainRect(mainRectX, mainRectY)

基本上，矩形从屏幕顶部开始，然后完全击中它的底部。正如预期的那样，它会向控制台输出很多“b”。然后，矩形向上移动一个像素，程序打印一个“a”到控制台，但然后它再次下降，即使“if not down down和rectOneY！= 0”表达式等于True和“if” rectOneY＆lt; height - 21 and goingDown“表达式等于False。

我一直试图解决这个问题至少一个小时，我根本无法理解什么是错的，我可以使用一些帮助。

我希望你指出我的代码有什么问题（只有我要问的具体问题，而不是xD之前的数百万个坏代码示例）。

提前致谢。

Answer 1

while running循环的第一行是goingDown = True。这会在循环的每次迭代开始时将变量goingDown设置为true。您需要在循环之前放置行goingDown = True。

它应该是这样的：

#GameLoop
goingDown = True
while running:
    ...

Python和Pygame - 尝试上下弹跳矩形

1 个答案: