Question

我做了一个小游戏，但问你可以猜，我有一个问题：p

我的第一堂课（它从游戏的第一部分获得原版画面）

    import pygame
    from pygame.locals import *
    import pygbutton

class part():
 def __init__(self,originalscreen):

    self.screen = originalscreen
    self.part()

 def part(self):

    """ Finally: the game!"""
    # But first, some variables and imports
    parton = True
    import makemap

    # Looping
    while parton:
        # The screen
        self.screen.fill(0)
        makemap.mapmaker(self.screen)

        # Events
        for event in pygame.event.get():
            if event.type == QUIT:
                exit()

        # Updating the screen
        pygame.display.flip()

正如您所读，我导入了makemap并将屏幕传递给它

   import pygame
   from pygame.locals import *

   """ Add tiles here"""
   grass = pygame.image.load("tiles/grass.png")

   def mapmaker(screen):
     height = screen.get_height()
     width = screen.get_width()

     # Set the size of each tile (25 pixels)
     xscale = width / 40
     yscale = height / 24

     # Loading file
     mapfile = open("part1map.txt", "r")

     # Making the map
     xco = 0
     yco = 0

     lines = mapfile.readlines()
     for line in lines:
        for typeoftile in range(0, len(line)):
            if str(typeoftile) == "g":
                screen.blit(grass, (xco*xscale, yco*yscale))
                xco = xco + 1
        yco = yco + 1

然后我的问题出现了：我的背景保持黑色（0），因此mapmaker功能不会将屏幕传回我的第一个代码文件。

如何编辑屏幕（就像我在mapmaker中那样）然后显示它？

我希望有人可以帮助我卢卡斯

ps：如果您有任何疑问，请询问。对不起我的英语，我是荷兰人......

Answer 1

您的问题尚未完全指定 - 例如，我们不知道地图文件的外观。但这部分肯定是错误的：

for line in lines:
    for typeoftile in range(0, len(line)):
        if str(typeoftile) == "g":
            screen.blit(grass, (xco*xscale, yco*yscale))
            xco = xco + 1

“typeoftile”将绑定到0和len（line）之间的数字 - 1.这显然永远不会是字符“g”，因此条件永远不会触发。< / p>

看起来你想要的是相应的字符，而不是索引。最简单的解决方法是插入索引操作符，如下所示：

for line in lines:
    for index in range(0, len(line)):
        typeoftile = line[index]
        if str(typeoftile) == "g":
            screen.blit(grass, (xco*xscale, yco*yscale))
            xco = xco + 1

然而，它更优雅，“Pythonic”迭代字符串中的字符，如下所示：

for line in lines:
    for typeoftile in line:
        if str(typeoftile) == "g":
            screen.blit(grass, (xco*xscale, yco*yscale))
            xco = xco + 1

假设“line”是一个字符串，那么“typeoftile”将依次绑定到字符串中的每个字符。（“abc”在迭代这样的字符时与[“a”，“b”，“c”]的作用相同。）

Answer 2

但是我仍然有一个问题：只显示了一条水平线的“草地”，所以它看起来像“yco = yco + 1”不起作用，但确实如此（我通过让我的程序测试了这一点）在for-loop中打印yco。

那么原因是什么原因不能显示第二（和第三......）水平草线？

代码现在看起来像这样：

    xco = 0
    yco = 0

    lines = mapfile.readlines()
    for line in lines:
        for typeoftile in line:
            if str(typeoftile) == "g":
                screen.blit(grass, (xco*xscale, yco*yscale))
                xco = xco + 1
        yco = yco + 1

    return screen

顺便说一句，输入文件如下所示：

gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg gggggggggggggggggggggggggggggggggggggggg

Pygame，改变屏幕和传回

2 个答案: