Countdown timer in Pygame

问题:

I started using pygame and I want to do simple game. One of the elements which I need is countdown timer. How can I do the countdown time (eg 10 seconds) in PyGame?

回答1:

On this page you will find what you are looking for http://www.pygame.org/docs/ref/time.html#pygame.time.get_ticks
You download ticks once before beginning the countdown (which can be a trigger in the game - the key event, whatever). For example:

start_ticks=pygame.time.get_ticks() #starter tick
while mainloop: # mainloop
    seconds=(pygame.time.get_ticks()-start_ticks)/1000 #calculate how many seconds
    if seconds>10: # if more than 10 seconds close the game
        break
    print (seconds) #print how many seconds

回答2:

Another easy way is to simply use pygame\'s event system.

Here\'s a simple example:

import pygame
pygame.init()
screen = pygame.display.set_mode((128, 128))
clock = pygame.time.Clock()

counter, text = 10, \'10\'.rjust(3)
pygame.time.set_timer(pygame.USEREVENT, 1000)
font = pygame.font.SysFont(\'Consolas\', 30)

while True:
    for e in pygame.event.get():
        if e.type == pygame.USEREVENT: 
            counter -= 1
            text = str(counter).rjust(3) if counter > 0 else \'boom!\'
        if e.type == pygame.QUIT: break
    else:
        screen.fill((255, 255, 255))
        screen.blit(font.render(text, True, (0, 0, 0)), (32, 48))
        pygame.display.flip()
        clock.tick(60)
        continue
    break

回答3:

pygame.time.Clock.tick returns the time in milliseconds since the last clock.tick call (delta time, dt), so you can use it to increase or decrease a timer variable.

import pygame as pg


def main():
    pg.init()
    screen = pg.display.set_mode((640, 480))
    font = pg.font.Font(None, 40)
    gray = pg.Color(\'gray19\')
    blue = pg.Color(\'dodgerblue\')
    # The clock is used to limit the frame rate
    # and returns the time since last tick.
    clock = pg.time.Clock()
    timer = 10  # Decrease this to count down.
    dt = 0  # Delta time (time since last tick).

    done = False
    while not done:
        for event in pg.event.get():
            if event.type == pg.QUIT:
                done = True

        timer -= dt
        if timer <= 0:
            timer = 10  # Reset it to 10 or do something else.

        screen.fill(gray)
        txt = font.render(str(round(timer, 2)), True, blue)
        screen.blit(txt, (70, 70))
        pg.display.flip()
        dt = clock.tick(30) / 1000  # / 1000 to convert to seconds.


if __name__ == \'__main__\':
    main()
    pg.quit()

回答4:

There are several ways you can do this- here\'s one. Python doesn\'t have a mechanism for interrupts as far as I know.

import time, datetime

timer_stop = datetime.datetime.utcnow() +datetime.timedelta(seconds=10)
while True:
    if datetime.datetime.utcnow() > timer_stop:
        print \"timer complete\"
        break

回答5:

This is actually quite simple. Thank Pygame for creating a simple library!

import pygame
x=0
while x < 10:
    x+=1
    pygame.time.delay(1000)

That\'s all there is to it! Have fun with pygame!