我正试图在Pygame中写下'破砖者'游戏。但我现在卡住了,不知道该怎么做。
(等等等等,等等,更清楚地解释情况,等等等等,明确,明确的情况)(我需要更多随机文字,所以我可以发布所有这些代码)
以下是所有代码,(是的全部):
import pygame, sys, time, random
from pygame.locals import *
pygame.init()
fpsclock = pygame.time.Clock()
WINDOWWIDTH = 450
WINDOWHEIGHT = 650
mainwindow = pygame.display.set_mode((WINDOWWIDTH, WINDOWHEIGHT), 0, 32)
pygame.display.set_caption('Luzion - Brick Breaker')
paddle = pygame.image.load('Brick Breaker - Paddle.png')
paddlerect = paddle.get_rect()
paddlerect.topleft = (190, 575)
ball = pygame.image.load ('ball.png')
ballrect = ball.get_rect()
ballrect.topleft = (195, 565)
cooltext = pygame.image.load('cooltext1.png')
cooltextrect = cooltext.get_rect()
cooltextrect.topleft = (0, 0)
BLACK = (0, 0, 0)
WHITE = (255, 255, 255)
RED = (255, 0, 0)
GREEN = (0, 128, 0)
BLUE = (0, 0, 255)
LIME = (0, 255, 0)
TEXTCOLOR = WHITE
font = pygame.font.SysFont(None, 48)
def displaytext(text, font, surface, x, y):
text = font.render(text, 1, TEXTCOLOR)
textrect = text.get_rect()
textrect.topleft = (x, y)
surface.blit(text, textrect)
def waitforplayer():
while True:
for event in pygame.event.get():
if event.type == QUIT:
pygame.quit()
sys.exit()
if event.type == KEYDOWN:
if event.key == K_ESCAPE:
pygame.quit()
sys.exit()
return
moveleft = False
moveright = False
SPEED = 7
bmoveup = bmovedown = bmoveleft = bmoveright = False
BALLSPEED = 8
mainwindow.blit(cooltext, cooltextrect)
pygame.display.update()
time.sleep(1)
displaytext('Level 1', font, mainwindow, 150, 100)
pygame.display.update()
time.sleep(1)
displaytext('Press any key to begin...', font, mainwindow, 22, 200)
pygame.display.update()
waitforplayer()
while True:
rb = pygame.image.load('redblock.png')
rbrect = rb.get_rect()
rbrect.topleft = (0, 0)
rb1 = rb
rb1rect = rb1.get_rect()
rb1rect.topleft = (40, 0)
level1blocks = [rb, rb1]
level1rects = [rbrect, rb1rect]
number = random.randint(0, 1)
for event in pygame.event.get():
if event.type == QUIT:
pygame.quit()
sys.exit()
if event.type == KEYDOWN:
if event.key == ord('a') or event.key == K_LEFT:
moveleft = True
moveright = False
if event.key == ord('d') or event.key == K_RIGHT:
moveleft = False
moveright = True
if event.key == ord('g'):
bmoveup = True
if number == 1:
bmoveleft = True
else:
bmoveright = True
if event.type == KEYUP:
if event.key == ord('a') or event.key == K_LEFT:
moveleft = False
if event.key == ord('d') or event.key == K_RIGHT:
moveright = False
if moveleft and paddlerect.left > 0:
paddlerect.left -= SPEED
if moveright and paddlerect.right < WINDOWWIDTH:
paddlerect.right += SPEED
if bmovedown and ballrect.bottom < WINDOWHEIGHT:
ballrect.top += BALLSPEED
if bmoveup and ballrect.top > 0:
ballrect.top -= BALLSPEED
if bmoveleft and ballrect.left > 0:
ballrect.left -= BALLSPEED
if bmoveright and ballrect.right < WINDOWWIDTH:
ballrect.right += BALLSPEED
if ballrect.top <= 0:
bmovedown = not bmovedown
bmoveup = not bmoveup
if ballrect.left <= 0:
bmoveleft = not bmoveleft
bmoveright = not bmoveright
if ballrect.right >= WINDOWWIDTH:
bmoveleft = not bmoveleft
bmoveright = not bmoveright
if ballrect.bottom >= WINDOWHEIGHT:
bmovedown = not bmovedown
bmoveup = not bmoveup
mainwindow.fill(WHITE)
mainwindow.blit(paddle, paddlerect)
mainwindow.blit(ball, ballrect)
for x in range(len(level1blocks)):
mainwindow.blit(level1blocks[x], level1rects[x])
for x in level1rects:
if ballrect.colliderect(x):
level1rects.remove([x])
level1blocks.remove([x])
if ballrect.colliderect(paddlerect):
bmovedown = not bmovedown
bmoveup = not bmoveup
bmoveleft = not bmoveleft
bmoveright = not bmoveright
pygame.display.update()
fpsclock.tick(35)
这是我的错误:
Traceback (most recent call last):
File "C:/Python32/Luzion - Brick Breaker", line 144, in <module>
level1rects.remove([x])
ValueError: list.remove(x): x not in list
请帮忙。
答案 0 :(得分:4)
正如你在评论中提到的,for循环都在一个较大的while循环中。这意味着当行
level1rects.remove(x)
发生,它会减少level1rects
的大小,导致level1rects[x]
(当x
为1时)引发异常。
解决此问题的最佳方法是将for循环更改为以下内容:
for b in level1blocks:
mainwindow.blit(b, b.get_rect())
for x in level1blocks[:]:
if ballrect.colliderect(x.get_rect()):
level1blocks.remove(x)
这省略了对level1rects
的需求 - 它造成了太多问题,因为你从矩形列表中删除了项目而没有从块列表中删除相应的块。将第一个循环更改为for b in level1blocks:
可以使代码在块消失时工作。
答案 1 :(得分:-1)
在level1blocks和level1rects列表上执行print或pprint,其中一个没有足够的成员用于range(2)迭代器,它将尝试访问第一个(索引0)和第二个(索引1)成员每个清单。