Question

我有一个使用python和pygame的代码，当满足某些条件时，它应该擦除屏幕并在中间打印“GAME OVER”字样。但是当条件得到满足时，与玩家发生碰撞的棋子就会直接通过并且不做任何事情。如何制作以便屏幕删除并打印出文字？

import pygame
import random
BLACK = (0,0,0)
WHITE = (255,255,255)
GREEN = (0,255,0)
RED = (255,0,0)
BLUE = (0,0,255)
pygame.init()
size = (700,700)
screen = pygame.display.set_mode(size)
pygame.display.set_caption("Dodger")
done = False
clock = pygame.time.Clock()

bullet_x = 350
bullet_y = 0

bullet_x2 = 175
bullet_y2 = 0

bullet_x3 = 525
bullet_y3 = 0

circle_x = 350
while not done:
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            done = True
        if event.type == pygame.KEYDOWN:
            circle_x += 5
    bullet_y += 1
    bullet_y2 += 2
    bullet_y3 += 3
    screen.fill(BLACK)
    pygame.draw.circle(screen, GREEN, (circle_x,600), 15)
    pygame.draw.circle(screen, RED, (bullet_x, bullet_y), 20)
    pygame.draw.circle(screen, RED, (bullet_x2, bullet_y2), 20)
    pygame.draw.circle(screen, RED, (bullet_x3, bullet_y3), 20)
    if bullet_y == 800:
        bullet_y = 0
        bullet_x = random.randint(20, 680)
    if bullet_y2 == 800:
        bullet_y2 = 0
        bullet_x2 = random.randint(20, 680)
    if bullet_y3 == 800:
        bullet_y3 = 0
        bullet_x3 = random.randint(20, 680)
    if circle_x == 685:
        circle_x = 15
    if bullet_y == 600 and bullet_x == circle_x or bullet_y2 == 600 and bullet_x2 == circle_x or bullet_y3 == 600 and bullet_x3 == circle_x:
        screen.fill(BLACK)
        font = pygame.font.SysFont('Calibri',40,True,False)
        text = font.render("GAME OVER",True,RED)
        screen.blit(text,[0,0])
    pygame.display.flip()
    clock.tick(300)
pygame.quit()

Answer 1

将变量circle_y设置为600（circle_y = 600），在circle_x = 350的正下方。然后将第50行更改为：

if (abs(bullet_y - circle_y) < 35 and abs(bullet_x - circle_x) < 35) or (abs(bullet_y2 - circle_y) < 35 and abs(bullet_x2 - circle_x) < 35) or abs(bullet_y3 - circle_y) < 35 and abs(bullet_x3 - circle_x) < 35:

这将解决您的碰撞问题。我可以建议你将这个表达重构为

do_circle_and_bullet_hit = abs(bullet_y - circle_y) < 35 and abs(bullet_x - circle_x) < 35
do_circle_and_bullet2_hit = abs(bullet_y2 - circle_y) < 35 and abs(bullet_x2 - circle_x) < 35
do_circle_and_bullet3_hit = abs(bullet_y3 - circle_y) < 35 and abs(bullet_x3 - circle_x) < 35
if do_circle_and_bullet_hit or do_circle_and_bullet2_hit or do_circle_and_bullet3_hit:

我们的想法是检查圆圈和每个子弹之间的距离。如果距离低于35，他们就会命中。

Answer 2

我认为创建Pygame Rect Class以测试碰撞会有所帮助。它们是通过bulletrect = pygame.rect.Rect(xpos, ypos, xsize, ysize)来完成的。稍后在代码中，您可以执行bulletrect.y += 1,2或3来使其失效。执行bulletrect.x =或bulletrect.y =会更改圈子的x和y位置。要在屏幕上显示圈子，请将pygame.draw.circle(screen, RED, (bullet_x, bullet_y), 20)或任何其他圈子更改为pygame.draw.circle(screen, RED, bulletrect.center, 20)。要实际检查冲突，请使用colliderect方法：playerrect.colliderect(bulletrect)。代码是：

import pygame
import random
BLACK = (0,0,0)
WHITE = (255,255,255)
GREEN = (0,255,0)
RED = (255,0,0)
BLUE = (0,0,255)
pygame.init()
size = (700,700)
screen = pygame.display.set_mode(size)
pygame.display.set_caption("Dodger")
done = False
clock = pygame.time.Clock()

bulletrect1 = pygame.rect.Rect((350, 0, 20, 20))
bulletrect2 = pygame.rect.Rect((175, 0, 20, 20))
bulletrect3 = pygame.rect.Rect((525, 0, 20, 20))

circlerect = pygame.rect.Rect((350, 600, 20, 20))
while not done:
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            done = True
        if event.type == pygame.KEYDOWN:
            circle.rect.x += 5
    bulletrect1.rect.y += 1
    bulletrect2.rect.y2 += 2
    bulletrect3.rect.y3 += 3
    screen.fill(BLACK)
    pygame.draw.circle(screen, GREEN, (circlerect.center), 15)
    pygame.draw.circle(screen, RED, (bulletrect1.center), 20)
    pygame.draw.circle(screen, RED, (bulletrect2.center), 20)
    pygame.draw.circle(screen, RED, (bulletrect3.center), 20)
    if bulletrect1.y == 800:
        bulletrect1.y = 0
        bulletrect1.x = random.randint(20, 680)
    if bulletrect2.y == 800:
        bulletrect2.y = 0
        bulletrect2.x = random.randint(20, 680)
    if bulletrect3.y == 800:
        bulletrect3.y = 0
        bulletrect3.x = random.randint(20, 680)
    if circlerect.x == 685:
        circlerect.x = 15
    if [bulletrect.colliderect(circlerect) for bulletrect in (bulletrect1, bulletrect2, bulletrect3)]
        screen.fill(BLACK)
        font = pygame.font.SysFont('Calibri',40,True,False)
        text = font.render("GAME OVER",True,RED)
        screen.blit(text,[0,0])
    pygame.display.flip()
    clock.tick(300)
pygame.quit()

另一个了解pygame和pygame rects的好地方是Al Sweigarts free online book教python。 Chapter 2教授rects。

编辑：

最后一段代码并没有真正起作用，所以这里有一些新的和改进的代码：

import pygame
import random
BLACK = (0,0,0)
WHITE = (255,255,255)
GREEN = (0,255,0)
RED = (255,0,0)
BLUE = (0,0,255)
pygame.init()
size = (700,700)
screen = pygame.display.set_mode(size)
pygame.display.set_caption("Dodger")
done = False
clock = pygame.time.Clock()

def resetpositions():
    global bulletrect1, bulletrect2, bulletrect3, circlerect
    bulletrect1 = pygame.rect.Rect((350, 0, 20, 20))
    bulletrect2 = pygame.rect.Rect((175, 0, 20, 20))
    bulletrect3 = pygame.rect.Rect((525, 0, 20, 20))

    circlerect = pygame.rect.Rect((350, 600, 20, 20))

resetpositions()
while not done:
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            done = True
        if event.type == pygame.KEYDOWN:
            circlerect.x += 5
    bulletrect1.y += 1
    bulletrect2.y += 2
    bulletrect3.y += 3
    screen.fill(BLACK)
    pygame.draw.circle(screen, GREEN, (circlerect.center), 15)
    pygame.draw.circle(screen, RED, (bulletrect1.center), 20)
    pygame.draw.circle(screen, RED, (bulletrect2.center), 20)
    pygame.draw.circle(screen, RED, (bulletrect3.center), 20)
    if bulletrect1.y == 800:
        bulletrect1.y = 0
        bulletrect1.x = random.randint(20, 680)
    if bulletrect2.y == 800:
        bulletrect2.y = 0
        bulletrect2.x = random.randint(20, 680)
    if bulletrect3.y == 800:
        bulletrect3.y = 0
        bulletrect3.x = random.randint(20, 680)
    if circlerect.x == 685:
        circlerect.x = 15
    if circlerect.collidelist((bulletrect1, bulletrect2, bulletrect3)) == 0:
        screen.fill(BLACK)
        font = pygame.font.SysFont('Calibri',40,True,False)
        text = font.render("GAME OVER",True,RED)
        screen.blit(text,[0,0])
        pygame.display.update()
        pygame.time.delay(3000)
        resetpositions()
    pygame.display.flip()
    clock.tick(300)

我改变了所以现在它在屏幕上显示GAME OVER时调用pygame.display.update（）。它也将等待三秒钟，因此GAME OVER不会立即消失。此外，它现在实际上工作。访问这两个网站，以清楚地了解pygame.rect.Rect的用途。它有很大的帮助。

如何让屏幕擦除然后在pygame中绘制新内容？

2 个答案: