Question

我几乎放弃了，所以我需要帮助。我正在尝试制作一个双跳的简单平台游戏...... 我一直在试图找到有用的东西到目前为止我最好的想法是比较滴答的数量，但每次我得到一些想法，我都会设法把它搞砸了，我不知道怎么样...... 有一种简单的方法吗？

请忽略任何不必要的变量，这只是一个例子

Clock=pygame.time.Clock()

t=0
a=0
b=0
f=0
m=0

while True:
    Clock.tick(180)
    for event in pygame.event.get():
        if event.type == QUIT:       
            pygame.quit()
            sys.exit()
        if event.type == KEYDOWN:
            if event.key==K_SPACE and b==0:
                movey=-1
                a=1
                t=pygame.time.get_ticks()
            if event.key==K_SPACE and b==1:
                f=pygame.time.get_ticks()
                if f<=t+238:
                    a=1
                else:
                    pass
        if event.type == KEYUP:
            if event.key==K_SPACE and b==0 and a==1:
                a=0
                b=1
            if event.key==K_SPACE and b==1 and a==1:
                m=1
                b=3
                s=y
    if m==1:
        y+=movey
        if y==s-32:
            m==0

    elif y<=312 and movey==-1:
        movey=+1
    elif y==344 and movey==+1:
        movey=0
        a=0
        b=0
    else:
        y+=movey

现在这是我正在尝试的......

Answer 1

g = -1  # gravity
floor = 0  # where frog stands


class Frog():  # say you have a frog
    def __init__(self):
        self.y = 0  # distance from ground
        self.y_speed = 0  # speed
        self.jumping = 0  # jumping status

    def jump(self):
        if self.jumping == 0:
            self.y_speed = 9  # a big jump
            self.jumping = 1  # change jumping status
        # I want the small jump available only when falling
        elif self.jumping == 1 and self.y_speed <= 0:
            self.y_speed = 5  # a small one
            self.jumping = 2  # change jumping status

    def update(self):  # this is called by mainloop
        self.y_speed += g  # change the acceleration
        self.y = max(self.y + self.y_speed, floor)  # don't want fall off
        if self.y == 0:  # on the ground again!
            self.jumping = 0  # reset jump
            self.y_speed = 0  # reset speed

Answer 2

两个例子：

单跳http://pastebin.com/ZLuPBg7k
双跳http://pastebin.com/uwK5wPM1

你可以强制双跳之间的最短时间，但我把它留下来简化示例。询问您是否还有其他问题。

Answer 3

使用柜台会不会更容易？只需在你的播放器中设置一个计数器变量= 0，然后定义你的跳转并设置计数器+ = 1.然后说如果计数器＆lt; 2然后你就可以跳了。不是吗？

Pygame双跳

3 个答案: