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
原文：https://blog.csdn.net/qq_42847252/article/details/83153696

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
   
   上代码

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<event.pos[0]<275 and 100<event.pos[1]<200:
						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<event.pos[0]<275 and 300<event.pos[1]<400:
						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<event.pos[0]<350 and 450<event.pos[1]<550:
						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<len(map1[0]) and map1[i-1][j+1] == '9':
		count += 1
	if j-1>=0 and map1[i][j-1] == '9':
		count += 1
	if j+1<len(map1[0]) and map1[i][j+1] == '9':
		count += 1
	if i+1<len(map1) and j-1>=0 and map1[i+1][j-1] == '9':
		count += 1
	if i+1<len(map1) and map1[i+1][j] == '9':
		count += 1
	if i+1<len(map1) and j+1<len(map1[0]) and map1[i+1][j+1] == '9':
		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<len(map1[0]) and len(map1[i-1][j+1]) == 0 and map2[i-1][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<len(map1[0]) and len(map1[i][j+1]) == 0 and map2[i][j+1] == '':
				arround(screen,i,j+1,map1,map2)
			if i+1<len(map1) 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<len(map1) and len(map1[i+1][j]) == 0 and map2[i+1][j] == '':
				arround(screen,i+1,j,map1,map2)
			if i+1<len(map1) and j+1<len(map1[0]) and len(map1[i+1][j+1]) == 0 and map2[i+1][j+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<event.pos[0]<350 and 450<event.pos[1]<550:
						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。