如何在Pygame中动态显示？

Question

我可以在显示屏上显示数字“ 1”。现在，我的目标是显示数字1 ... 4。这意味着首先显示“ 1”，然后显示1“ 2”，依此类推。如何解决该问题？

这是我的代码：

import pygame
import time

pygame.init()
screen = pygame.display.set_mode((640, 480))
clock = pygame.time.Clock()
done = False

font = pygame.font.SysFont("comicsansms", 72)  
text = font.render("1", True, (0, 128, 0))

while not done:
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            done = True
        if event.type == pygame.KEYDOWN and event.key == pygame.K_ESCAPE:
            done = True

    screen.fill((255, 255, 255))
    screen.blit(text,
                (320 - text.get_width() // 2, 240 - text.get_height() // 2))

    pygame.display.flip()
    clock.tick(60)

Answer 1

创建一个计数器，然后我们用str()将值转换为字符串：

counter = 1
text = font.render(str(counter), True, (0, 128, 0))

以一定的时间间隔添加计时器事件（请参见pygame.event）。计时器事件由pygame.time.set_timer()

启动

mytimerevent = pygame.USEREVENT + 1
pygame.time.set_timer(mytimerevent, 1000) # 1000 milliseconds = 1 socond 

增加计数器并更改事件上的文字：

while not done:
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            done = True
        if event.type == pygame.KEYDOWN and event.key == pygame.K_ESCAPE:
            done = True

        if event.type == mytimerevent: # timer event
            counter += 1
            text = font.render(str(counter), True, (0, 128, 0))


    # [...]

Answer 2

每帧更新text如下：

# Setup stuff

number = 1
text = font.render(str(number), True, (0, 128, 0))

while not done:
    # Show stuff on screen

    number += 1
    text = font.render(str(number), True, (0, 128, 0))