自己写一个 Python 消消乐小游戏吧!

裸睡的猪

共 6067字,需浏览 13分钟

 ·

2020-07-31 16:34

文 | 野客

来源:Python 技术「ID: pythonall」


提到开心消消乐这款小游戏,相信大家都不陌生,其曾在 2015 年获得过玩家最喜爱的移动单机游戏奖,受欢迎程度可见一斑,本文我们使用 Python 来做个简单的消消乐小游戏。

实现

消消乐的构成主要包括三部分:游戏主体、计分器、计时器,下面来看一下具体实现。

先来看一下游戏所需 Python 库。

import osimport sysimport timeimport pygameimport random

定义一些常量,比如:窗口宽高、网格行列数等,代码如下:

WIDTH = 400HEIGHT = 400NUMGRID = 8GRIDSIZE = 36XMARGIN = (WIDTH - GRIDSIZE * NUMGRID) // 2YMARGIN = (HEIGHT - GRIDSIZE * NUMGRID) // 2ROOTDIR = os.getcwd()FPS = 30

接着创建一个主窗口,代码如下:

pygame.init()screen = pygame.display.set_mode((WIDTH, HEIGHT))pygame.display.set_caption('消消乐')

看一下效果:

再接着在窗口中画一个 8 x 8 的网格,代码如下:

screen.fill((255, 255, 220))# 游戏界面的网格绘制def drawGrids(self):	for x in range(NUMGRID):		for y in range(NUMGRID):			rect = pygame.Rect((XMARGIN+x*GRIDSIZE, YMARGIN+y*GRIDSIZE, GRIDSIZE, GRIDSIZE))			self.drawBlock(rect, color=(255, 165, 0), size=1# 画矩形 block 框def drawBlock(self, block, color=(255, 0, 0), size=2):	pygame.draw.rect(self.screen, color, block, size)

看一下效果:

再接着在网格中随机放入各种拼图块,代码如下:

while True:	self.all_gems = []	self.gems_group = pygame.sprite.Group()	for x in range(NUMGRID):		self.all_gems.append([])		for y in range(NUMGRID):			gem = Puzzle(img_path=random.choice(self.gem_imgs), size=(GRIDSIZE, GRIDSIZE), position=[XMARGIN+x*GRIDSIZE, YMARGIN+y*GRIDSIZE-NUMGRID*GRIDSIZE], downlen=NUMGRID*GRIDSIZE)			self.all_gems[x].append(gem)			self.gems_group.add(gem)	if self.isMatch()[0] == 0:		break

看一下效果:

再接着加入计分器和计时器,代码如下:

# 显示得分def drawScore(self):	score_render = self.font.render('分数:'+str(self.score), 1, (85, 65, 0))	rect = score_render.get_rect()	rect.left, rect.top = (55, 15)	self.screen.blit(score_render, rect)# 显示加分def drawAddScore(self, add_score):	score_render = self.font.render('+'+str(add_score), 1, (255, 100, 100))	rect = score_render.get_rect()	rect.left, rect.top = (250, 250)	self.screen.blit(score_render, rect)# 显示剩余时间def showRemainingTime(self):	remaining_time_render = self.font.render('倒计时: %ss' % str(self.remaining_time), 1, (85, 65, 0))	rect = remaining_time_render.get_rect()	rect.left, rect.top = (WIDTH-190, 15)	self.screen.blit(remaining_time_render, rect)

看一下效果:

当设置的游戏时间用尽时,我们可以生成一些提示信息,代码如下:

while True:	for event in pygame.event.get():		if event.type == pygame.QUIT:			pygame.quit()			sys.exit()		if event.type == pygame.KEYUP and event.key == pygame.K_r:			flag = True	if flag:		break	screen.fill((255, 255, 220))	text0 = '最终得分: %s' % score	text1 = '按 R 键重新开始'	y = 140	for idx, text in enumerate([text0, text1]):		text_render = font.render(text, 1, (85, 65, 0))		rect = text_render.get_rect()		if idx == 0:			rect.left, rect.top = (100, y)		elif idx == 1:			rect.left, rect.top = (100, y)		y += 60		screen.blit(text_render, rect)	pygame.display.update()

看一下效果:

说完了游戏图形化界面相关的部分,我们再看一下游戏的主要处理逻辑。

我们通过鼠标来操纵拼图块,因此程序需要检查有无拼图块被选中,代码实现如下:

def checkSelected(self, position):	for x in range(NUMGRID):		for y in range(NUMGRID):			if self.getGemByPos(x, y).rect.collidepoint(*position):				return [x, y]	return None

我们需要将鼠标连续选择的拼图块进行位置交换,代码实现如下:

def swapGem(self, gem1_pos, gem2_pos):	margin = gem1_pos[0] - gem2_pos[0] + gem1_pos[1] - gem2_pos[1]	if abs(margin) != 1:		return False	gem1 = self.getGemByPos(*gem1_pos)	gem2 = self.getGemByPos(*gem2_pos)	if gem1_pos[0] - gem2_pos[0] == 1:		gem1.direction = 'left'		gem2.direction = 'right'	elif gem1_pos[0] - gem2_pos[0] == -1:		gem2.direction = 'left'		gem1.direction = 'right'	elif gem1_pos[1] - gem2_pos[1] == 1:		gem1.direction = 'up'		gem2.direction = 'down'	elif gem1_pos[1] - gem2_pos[1] == -1:		gem2.direction = 'up'		gem1.direction = 'down'	gem1.target_x = gem2.rect.left	gem1.target_y = gem2.rect.top	gem1.fixed = False	gem2.target_x = gem1.rect.left	gem2.target_y = gem1.rect.top	gem2.fixed = False	self.all_gems[gem2_pos[0]][gem2_pos[1]] = gem1	self.all_gems[gem1_pos[0]][gem1_pos[1]] = gem2	return True

每一次交换拼图块时,我们需要判断是否有连续一样的三个及以上拼图块,代码实现如下:

def isMatch(self):	for x in range(NUMGRID):		for y in range(NUMGRID):			if x + 2 < NUMGRID:				if self.getGemByPos(x, y).type == self.getGemByPos(x+1, y).type == self.getGemByPos(x+2, y).type:					return [1, x, y]			if y + 2 < NUMGRID:				if self.getGemByPos(x, y).type == self.getGemByPos(x, y+1).type == self.getGemByPos(x, y+2).type:					return [2, x, y]	return [0, x, y]

当出现三个及以上拼图块时,需要将这些拼图块消除,代码实现如下:

def removeMatched(self, res_match):	if res_match[0] > 0:		self.generateNewGems(res_match)		self.score += self.reward		return self.reward	return 0

将匹配的拼图块消除之后,我们还需要随机生成新的拼图块,代码实现如下:

def generateNewGems(self, res_match):	if res_match[0] == 1:		start = res_match[2]		while start > -2:			for each in [res_match[1], res_match[1]+1, res_match[1]+2]:				gem = self.getGemByPos(*[each, start])				if start == res_match[2]:					self.gems_group.remove(gem)					self.all_gems[each][start] = None				elif start >= 0:					gem.target_y += GRIDSIZE					gem.fixed = False					gem.direction = 'down'					self.all_gems[each][start+1] = gem				else:					gem = Puzzle(img_path=random.choice(self.gem_imgs), size=(GRIDSIZE, GRIDSIZE), position=[XMARGIN+each*GRIDSIZE, YMARGIN-GRIDSIZE], downlen=GRIDSIZE)					self.gems_group.add(gem)					self.all_gems[each][start+1] = gem			start -= 1	elif res_match[0] == 2:		start = res_match[2]		while start > -4:			if start == res_match[2]:				for each in range(0, 3):					gem = self.getGemByPos(*[res_match[1], start+each])					self.gems_group.remove(gem)					self.all_gems[res_match[1]][start+each] = None			elif start >= 0:				gem = self.getGemByPos(*[res_match[1], start])				gem.target_y += GRIDSIZE * 3				gem.fixed = False				gem.direction = 'down'				self.all_gems[res_match[1]][start+3] = gem			else:				gem = Puzzle(img_path=random.choice(self.gem_imgs), size=(GRIDSIZE, GRIDSIZE), position=[XMARGIN+res_match[1]*GRIDSIZE, YMARGIN+start*GRIDSIZE], downlen=GRIDSIZE*3)				self.gems_group.add(gem)				self.all_gems[res_match[1]][start+3] = gem			start -= 1

之后反复执行这个过程,直至耗尽游戏时间,游戏结束。

最后,我们动态看一下游戏效果。

总结

本文我们使用 Python 实现了一个简单的消消乐游戏,有兴趣的可以对游戏做进一步扩展,比如增加关卡等。

源码:https://github.com/JustDoPython/python-examples/tree/master/yeke/py-xxl

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