**其实扫雷说是使用pygame写游戏,但其实这非常锻炼思维,一个编程与解决问题的思路的养成非常重要,这篇文章的主要内容不是讲解功能的实现方式,而是介绍我遇到的一些问题**

我自己的界面设计的能力有点差,所以在界面设计部分我借鉴了《python写扫雷小游戏(pygame)》—— 在校学渣一枚这位朋友的,再次表示十分感谢

好了,闲言少叙,一部分一部分的上

这个扫雷的主要实现思路就是用pygame划出格子后,界面上格子的行数列数,与另外一个二维列表的对应关系。

1. 难度选择界面

import pygame
import sys
import main_game
from pygame.locals import *
#初始化模块
pygame.init()
pygame.mixer.init()
#音效加载
button_sound = pygame.mixer.Sound('material/sound/button.wav')
button_sound.set_volume(0.2)
background_sound = pygame.mixer.music.load('material/sound/background.ogg')
pygame.mixer.music.set_volume(0.3)
sys.setrecursionlimit(2000)
#画出主界面
def draw_choose_interface(level):
#level为难度等级,prompt为显示界面所显示的文字
bg_size = width,height = 450,700
screen = pygame.display.set_mode(bg_size)
screen.fill((237,237,237))
#画出选择框
pygame.draw.rect(screen,(0,0,0),[175,100,100,300],5)
#分割出三部分存放+,-和等级的位置
pygame.draw.line(screen,(0,0,0),(175,200),(275,200),5)
pygame.draw.line(screen,(0,0,0),(175,300),(275,300),5)
#画出+的横线以及-
pygame.draw.line(screen,(0,0,0),(195,150),(255,150),20)
pygame.draw.line(screen,(0,0,0),(195,350),(255,350),20)
#画出+的竖线
pygame.draw.line(screen,(0,0,0),(225,120),(225,180),20)
#开始游戏的选择框
pygame.draw.rect(screen,(0,0,0),[100,450,250,100],5)
#定义字体跟大小
s_font1=pygame.font.Font('material/benmoyouyuan.ttf',50)
s_font2=pygame.font.Font('material/benmoyouyuan.ttf',16)
s_font3=pygame.font.Font('material/benmoyouyuan.ttf',34)
#文本确定
s_text1=s_font1.render(str(level),True,(0,0,0))
s_text2=s_font1.render("开始游戏",True,(0,0,0))
s_text3=s_font2.render('Listening_Rift',True,(0,0,0))
s_text4=s_font3.render('难度选择:',True,(255,0,0))
#将字放在窗口指定位置
screen.blit(s_text1,(200,220))
screen.blit(s_text2,(120,470))
screen.blit(s_text3,(22,650))
screen.blit(s_text4,(100,50))
pygame.display.set_caption('难度选择')
pygame.display.flip()
def _interface():
level = '低'
pygame.mixer.music.play(-1)
draw_choose_interface(level)
while True:
for event in pygame.event.get():
if event.type == QUIT:
pygame.quit()
sys.exit()
elif event.type == MOUSEBUTTONDOWN:
if event.button == 1:
if 175
if level == '低':
level = '中'
button_sound.play()
draw_choose_interface(level)
elif level == '中':
level = '高'
button_sound.play()
draw_choose_interface(level)
elif level == '高':
level = '低'
button_sound.play()
draw_choose_interface(level)
elif 175
if level == '高':
level = '中'
button_sound.play()
draw_choose_interface(level)
elif level == '中':
level = '低'
button_sound.play()
draw_choose_interface(level)
elif level == '低':
level = '高'
button_sound.play()
draw_choose_interface(level)
elif 100
button_sound.play()
main_game.game_main(level)
if __name__ == '__main__':
_interface()

这部分其实就主要是界面的绘制以及事件的确定,本身没有什么难度,都是一些简单的用法,整体在不到100行,这也是扫雷游戏的登入部分。

2. 接下来就要将整个游戏分块考虑并且编写,首先我把布雷部分写出来,这部分并不是很难,甚至可以说是最简单的,我是个初学者,五六十行就可以实现。

import random
def lay_mines(level,map1):
if level == '低':
a = 0
while a<10:
x = random.randint(0,9)
y = random.randint(0,9)
map1[x][y] = '9'
print('布雷功能运行正常')
a = 0
for i in range(10):
for j in range(10):
if map1[i][j] == '9':
a += 1
print(a)
elif level == '中':
a = 0
while a<40:
x = random.randint(0,15)
y = random.randint(0,15)
map1[x][y] = '9'
print('布雷功能运行正常')
a = 0
for i in range(16):
for j in range(16):
if map1[i][j] == '9':
a += 1
print(a)
elif level == '高':
a = 0
while a<99:
x = random.randint(0,31)
y = random.randint(0,15)
map1[x][y] = '9'
print('布雷功能运行正常')
a = 0
for i in range(32):
for j in range(16):
if map1[i][j] == '9':
a += 1
print(a)

布雷的部分实际上就是分等级并排出不同数量的雷

这里我遇到最大的问题就是“a = 0” 这一句,这里是想通过一个遍历的方法,数出十个雷,确保没有重复,但实际上random好像是有这个函数的,但是也有一定bug,还是要通过一些方法规避,如果没有这一句的话,a实际上是一个类加量,并不会真正布十个雷,初级4个,中级9个,高级14个,所以在计算之前一定要将其初始化为0

3. 点开事件

在扫雷游戏中,我们在点到一个非雷格子的时候,有的会出现数字有的会直接点开一片,在明确这个规则之后,我们就可以开始编写了

import pygame
import main_game
from pygame.locals import *
pygame.mixer.init()
boom_sound = pygame.mixer.Sound('material/sound/BOOM.wav')
boom_sound.set_volume(0.2)
#比较得出数字
def _digital(i,j,map1):
count = 0
if i-1>=0 and j-1>=0 and map1[i-1][j-1] == '9':
count += 1
if i-1>=0 and map1[i-1][j] == '9':
count += 1
if i-1>=0 and j+1
count += 1
if j-1>=0 and map1[i][j-1] == '9':
count += 1
if j+1
count += 1
if i+1=0 and map1[i+1][j-1] == '9':
count += 1
if i+1
count += 1
if i+1
count += 1
return count
#周围判断
def arround(screen,i,j,map1,map2):
if map1[i][j] == '9':
omine_image = pygame.image.load('material/picture/mine.gif').convert()
mine_image = pygame.transform.scale(omine_image,(22,22))
screen.blit(mine_image,(i*30+3,j*30+3))
boom_sound.play()
pygame.display.flip()
for x in range(len(map1)):
for y in range(len(map1[0])):
if map1[x][y] == '9':
screen.blit(mine_image,(x*30+3,y*30+3))
pygame.display.flip()
pygame.time.delay(100)
result = '游戏失败'
pygame.time.delay(1000)
main_game.result_screen(result)
#显示lose
else:
digital = _digital(i,j,map1)
map1[i][j] = str(digital)
print('内容填写部分正常')
if map1[i][j] == '0':
print('判断成功')
map1[i][j] = ' '
if i-1>=0 and j-1>=0 and len(map1[i-1][j-1]) == 0 and map2[i-1][j-1] == '':
arround(screen,i-1,j-1,map1,map2)
if i-1>=0 and len(map1[i-1][j]) == 0 and map2[i-1][j] == '':
arround(screen,i-1,j,map1,map2)
if i-1>=0 and j+1
arround(screen,i-1,j+1,map1,map2)
if j-1>=0 and len(map1[i][j-1]) == 0 and map2[i][j-1] == '':
arround(screen,i,j-1,map1,map2)
if j+1
arround(screen,i,j+1,map1,map2)
if i+1=0 and len(map1[i+1][j-1]) == 0 and map2[i+1][j-1] == '':
arround(screen,i+1,j-1,map1,map2)
if i+1
arround(screen,i+1,j,map1,map2)
if i+1
arround(screen,i+1,j+1,map1,map2)
s_font = pygame.font.Font('material/benmoyouyuan.ttf',19)
if map1[i][j] == '1':
color = (86,98,166)
elif map1[i][j] == '2':
color = (67,106,62)
elif map1[i][j] == '3':
color = (15,170,209)
else:
color = (222,29,90)
s_text = s_font.render(str(map1[i][j]),True,color)
#将字放在窗口指定位置
screen.blit(s_text,(i*30+9,j*30+3))
#将为零的格子填充为新颜色
for i in range(len(map1)):
for j in range(len(map1[0])):
if map1[i][j] == ' ':
pygame.draw.rect(screen,(200,200,200),[i*30,j*30,29,29])

靠通过审查周围雷的数量填入数字,如果为零就计算他周围的进行循环计算,这样就可以实现点开一大片。

问题:

- [ ] a、最初,当计算周围的时候在周围的两个空格子无限死循环下去。**解决方法:将为零的格子转换为空格,只计算内部内容有长度的格子。**

- [ ] b、但是,之后发现int变量是没有长度的,**就只好将所有的内容全部变为字符串,原本用9表示雷,现在则用'9'表示。**

这个部分实际上最重要的就是判断标准的确定,需要仔细地进行思考。

4. 主体游戏

好吧,我承认我的架构思路确实略有问题,原本是想把不同部分分开的后来在调试优化过程中,就逐渐的加到一起了。。。

import pygame
import traceback
import sys
import random
import choose
import lay_mines
import open_event
from pygame.locals import *
#9代表雷
#10代表旗帜
sys.setrecursionlimit(1000)
#板块初始化
pygame.init()
#音效载入,背景音乐,点击音效,爆炸音效,音量均为0.2
pygame.mixer.init()
button_sound = pygame.mixer.Sound('material/sound/button.wav')
button_sound.set_volume(0.2)
boom_sound = pygame.mixer.Sound('material/sound/BOOM.wav')
boom_sound.set_volume(0.2)
background_sound = pygame.mixer.music.load('material/sound/background.ogg')
pygame.mixer.music.set_volume(0.2)
#判定结果
def result_judge(map1,map2):
mine = []
flag = []
for x in range(len(map1)):
for y in range(len(map1[0])):
if map1[x][y] == '9':
mine.append((x,y))
if map2[x][y] == '10':
flag.append((x,y))
if mine == flag:
result = '游戏胜利'
else:
result = '游戏失败'
return result
def result_screen(result):
#建立界面
bg_size = width,height = 450,700
r_screen = pygame.display.set_mode(bg_size)
r_screen.fill((237,237,237))
r_font1 = pygame.font.Font('material/benmoyouyuan.ttf',67)
r_font2 = pygame.font.Font('material/benmoyouyuan.ttf',50)
pygame.draw.rect(r_screen,(0,0,0),[100,450,250,100],5)
r_text1 = r_font1.render(str(result),True,(0,0,0))
r_text2 = r_font2.render("继续游戏",True,(0,0,0))
r_screen.blit(r_text1,(90,100))
r_screen.blit(r_text2,(120,470))
pygame.display.set_caption('游戏结束')
pygame.display.flip()
while True:
for event in pygame.event.get():
if event.type == QUIT:
pygame.quit()
sys.exit()
elif event.type == MOUSEBUTTONDOWN:
if event.button == 1:
if 100
pygame.display.quit()
choose._interface()
def game_main(level):
#生成主界面
if level == '低':
bg_size = width,height = 300,300
elif level == '中':
bg_size = width,height = 480,480
elif level == '高':
bg_size = width,height = 960,480
screen_main = pygame.display.set_mode(bg_size)
screen_main.fill((237,237,237))
pygame.display.set_caption('L_R扫雷:%s级模式'%level)
pygame.mixer.music.play(-1)
#画格子
for x in range(width//30):
for y in range(height//30):
pygame.draw.rect(screen_main,(0,0,0),[x*30,y*30,29,29],1)
pygame.display.flip()
#初始化地图二维列表
if level == '低':
map1 = [[0 for col in range(10)] for row in range(10)]
map2 = [[0 for col in range(10)] for row in range(10)]
for i in range(10):
for j in range(10):
map1[i][j] = ''
map2[i][j] = ''
elif level == '中':
map1 = [[0 for col in range(16)] for row in range(16)]
map2 = [[0 for col in range(16)] for row in range(16)]
for i in range(16):
for j in range(16):
map1[i][j] = ''
map2[i][j] = ''
elif level == '高':
map1 = [[0 for col in range(16)] for row in range(32)]
map2 = [[0 for col in range(16)] for row in range(32)]
for i in range(32):
for j in range(16):
map1[i][j] = ''
map2[i][j] = ''
#布雷
lay_mines.lay_mines(level,map1)
while True:
for event in pygame.event.get():
if event.type == QUIT:
pygame.quit()
sys.exit()
elif event.type == MOUSEBUTTONDOWN:
if event.button == 1:
button_sound.play()
i = event.pos[0]//30
j = event.pos[1]//30
if (map1[i][j] == '' or map1[i][j] == '9') and map2[i][j] == '':
open_event.arround(screen_main,i,j,map1,map2)
print('翻开成功')
pygame.display.flip()
elif event.button == 3:
button_sound.play()
i = event.pos[0]//30
j = event.pos[1]//30
if (map1[i][j] == '' or map1[i][j] == '9') and map2[i][j] == '':
map2[i][j] = '10'
oflag = pygame.image.load('material/picture/flag.gif')
flag = pygame.transform.scale(oflag,(22,22))
screen_main.blit(flag,(i*30+3,j*30+3))
print('标记成功')
pygame.display.flip()
elif (map1[i][j] == '' or map1[i][j] == '9') and map2[i][j] == '10':
map2[i][j] = ''
print('取消标记')
oblank = pygame.image.load('material/picture/blank.gif')
flag = pygame.transform.scale(oblank,(22,22))
screen_main.blit(flag,(i*30+3,j*30+3))
pygame.display.flip()
if level == '低':
mines = 10
elif level == '中':
mines = 40
elif level == '高':
mines = 99
flags = 0
for x in range(len(map2)):
for y in range(len(map2[0])):
if map2[x][y] == '10':
flags += 1
if flags == mines:
result = result_judge(map1,map2)
pygame.time.delay(1000)
result_screen(result)

游戏主体部分包含胜负判断

最大的问题在于、标记时会影响判断,最一开始的标记是把二维列表中的值改编为10就会影响每个格子里的值,所以后来我又建立了另一张二维列表,专门存放旗子,这样就可以将他标记完成并不影响值。

最终还想实现战绩记录,每个模式的最短时间,以及胜率。这个部分连上界面优化就比较简单了,所以就日后慢慢进行吧。