更改 pygame window 标题栏的颜色

Change the color of a pygame window title bar

我目前正在使用 Python 3.7.0 编写贪吃蛇游戏。我使用 pygame。我是 Python 的新手,我不知道如何更改颜色,因为背景本身是黑色的,而且它上面的“Window 栏”也是黑色的。这使得很难看出字段在哪里结束。我不确定是否有可能这样做,但如果有人知道方法,请告诉我。对于图片我改变了背景颜色所以你可以明白我的意思This is the Game. Usually it has a Black background.

这是我目前使用的代码:

import pygame
import time
import random
 
pygame.init()
 
white = (255, 255, 255)
yellow = (255, 255, 102)
black = (0, 0, 0)
red = (213, 50, 80)
green = (0, 255, 0)
blue = (50, 153, 213)
 
dis_width = 600
dis_height = 400
 
dis = pygame.display.set_mode((dis_width, dis_height))
pygame.display.set_caption('Snake')
 
clock = pygame.time.Clock()
 
snake_block = 10
snake_speed = 15
 
font_style = pygame.font.SysFont("8514oem.fon", 40)
score_font = pygame.font.SysFont("8514oem.fon", 30)
 
def our_snake(snake_block, snake_list):
    for x in snake_list:
        pygame.draw.rect(dis, white, [x[0], x[1], snake_block, snake_block])
 
 
def message(msg, color):
    mesg = font_style.render(msg, True, color)
    dis.blit(mesg, [dis_width / 6, dis_height / 3])
 
 
def gameLoop():
    game_over = False
    game_close = False
 
    x1 = dis_width / 2
    y1 = dis_height / 2
 
    x1_change = 0
    y1_change = 0
 
    snake_List = []
    Length_of_snake = 1
 
    foodx = round(random.randrange(0, dis_width - snake_block) / 10.0) * 10.0
    foody = round(random.randrange(0, dis_height - snake_block) / 10.0) * 10.0
 
    while not game_over:
 
        while game_close == True:
            dis.fill(red)
            message("GAME OVER", white)
 
            pygame.display.update()
 
            for event in pygame.event.get():
                if event.type == pygame.KEYDOWN:
                    if event.key == pygame.K_ESCAPE:
                        game_over = True
                        game_close = False
                    if event.key == pygame.K_e:
                        game_over = True
                        game_close = False
                    if event.key == pygame.K_r:
                        gameLoop()
                    if event.key == pygame.K_SPACE:
                        gameLoop()
                    if event.key == pygame.K_RETURN:
                        gameLoop() 
        for event in pygame.event.get():
            if event.type == pygame.QUIT:
                game_over = True
            if event.type == pygame.KEYDOWN:
                if event.key == pygame.K_LEFT:
                    x1_change = -snake_block
                    y1_change = 0
                elif event.key == pygame.K_RIGHT:
                    x1_change = snake_block
                    y1_change = 0
                elif event.key == pygame.K_UP:
                    y1_change = -snake_block
                    x1_change = 0
                elif event.key == pygame.K_DOWN:
                    y1_change = snake_block
                    x1_change = 0
 
        if x1 >= dis_width or x1 < 0 or y1 >= dis_height or y1 < 0:
            game_close = True
        x1 += x1_change
        y1 += y1_change
        dis.fill(red)
        pygame.draw.rect(dis, green, [foodx, foody, snake_block, snake_block])
        snake_Head = []
        snake_Head.append(x1)
        snake_Head.append(y1)
        snake_List.append(snake_Head)
        if len(snake_List) > Length_of_snake:
            del snake_List[0]
 
        for x in snake_List[:-1]:
            if x == snake_Head:
                game_close = True
 
        our_snake(snake_block, snake_List)
 
 
        pygame.display.update()
 
        if x1 == foodx and y1 == foody:
            foodx = round(random.randrange(0, dis_width - snake_block) / 10.0) * 10.0
            foody = round(random.randrange(0, dis_height - snake_block) / 10.0) * 10.0
            Length_of_snake += 1
 
        clock.tick(snake_speed)
 
    pygame.quit()
    quit()
 
 
gameLoop()

代码来自 https://www.edureka.co/blog/snake-game-with-pygame/#install 我刚刚更改了它,所以它更适合我。

最好的问候 -桑德罗

我自己 pygame 并没有用太多,但我记得阅读以下内容可能会有所帮助

color = input("Choose a background color: ")
if color == "red":
    screen.fill(red)
    pygame.display.update()

不能 100% 确定这是否正确