Question

我想扩展这段代码，并使它对像我这样的初学者更基础。我对python非常陌生，不知道步骤是什么。我想使用更多代码，例如随机，范围，列表等。

我试图将xs和ys放入列表中，但是失败了。

import pygame, random, sys
from pygame.locals import *
def collide(x1, x2, y1, y2, w1, w2, h1, h2):
    if x1+w1>x2 and x1<x2+w2 and y1+h1>y2 and y1<y2+h2:return True
    else:return False
def die(screen, score):
    f=pygame.font.SysFont('Arial', 30);t=f.render('Your score was: '+str(score), True, (0, 0, 0));screen.blit(t, (10, 270));pygame.display.update();pygame.time.wait(2000);sys.exit(0)
xs = [290, 290, 290, 290, 290];ys = [290, 270, 250, 230, 210];dirs = 0;score = 0;applepos = (random.randint(0, 590), random.randint(0, 590));pygame.init();s=pygame.display.set_mode((600, 600));pygame.display.set_caption('Snake');appleimage = pygame.Surface((10, 10));appleimage.fill((0, 255, 0));img = pygame.Surface((20, 20));img.fill((255, 0, 0));f = pygame.font.SysFont('Arial', 20);clock = pygame.time.Clock()
while True:
    clock.tick(5)
    for e in pygame.event.get():
        if e.type == QUIT:
            sys.exit(0)
        elif e.type == KEYDOWN:
            if e.key == K_UP and dirs != 0:dirs = 2
            elif e.key == K_DOWN and dirs != 2:dirs = 0
            elif e.key == K_LEFT and dirs != 1:dirs = 3
            elif e.key == K_RIGHT and dirs != 3:dirs = 1
    i = len(xs)-1
    while i >= 2:
        if collide(xs[0], xs[i], ys[0], ys[i], 20, 20, 20, 20):die(s, score)
        i-= 1
    if collide(xs[0], applepos[0], ys[0], applepos[1], 20, 10, 20, 10):score+=1;xs.append(700);ys.append(700);applepos=(random.randint(0,590),random.randint(0,590))
    if xs[0] < 0 or xs[0] > 580 or ys[0] < 0 or ys[0] > 580: die(s, score)
    i = len(xs)-1
    while i >= 1:
        xs[i] = xs[i-1];ys[i] = ys[i-1];i -= 1
    if dirs==0:ys[0] += 20
    elif dirs==1:xs[0] += 20
    elif dirs==2:ys[0] -= 20
    elif dirs==3:xs[0] -= 20    
    s.fill((255, 255, 255))    
    for i in range(0, len(xs)):
        s.blit(img, (xs[i], ys[i]))
    s.blit(appleimage, applepos);t=f.render(str(score), True, (0, 0, 0));s.blit(t, (10, 10));pygame.display.update()

老实说，我不关心代码是否不像现在这样平滑，只要我能理解它就可以了。

35线蛇游戏

0 个答案: