pygame - KEYDOWN事件 - 不更新位置的行

时间:2013-01-29 10:37:21

标签: python line pygame keydown

我正在制作一个2人游戏,看起来应该是这样的:
Game image
在游戏中,射手(绿色和蓝色,由玩家控制)可以互相射击子弹 如果子弹
1.与墙壁碰撞(灰色),它被摧毁 2.击中射手,失去健康(射手) 游戏(假设是)基于回合制,并在玩家达到0健康状态时结束。

我的问题
我枪手的镜筒没有更新/旋转 2.有没有更好的方法来检测是否按下了(ny)键 *射手,子弹,墙是班级

我的代码
(如果没有提到任何有助于回答的功能,请发表评论)

import math,random,pygame
def event_handle(event,turn):
    if turn == 1:
        c_s = p1
    elif turn == 2:
        c_s = p2
    else:
        return None
    if event.type == pygame.KEYDOWN:
        key = pygame.key.get_pressed()
        # next_pos
        if key[pygame.K_q]:
            c_s.next_x -= 1
        if key[pygame.K_e]:
            c_s.next_x += 1
        # angle
        if key[pygame.K_w]:
            c_s.angle += radians(1)
        if key[pygame.K_s]:
            c_s.angle -= radians(1)
        # power (speed)
        if key[pygame.K_d]:
            c_s.speed += 0.1
        if key[pygame.K_a]:
            c_s.speed -= 0.1

def draw_all(bullist,shooters,wall,surface):
    # draw shooters
    for shooter in shooters:
        shooter.move()
        shooter.draw(surface)
    # draw bullets
    for bullet in bullist:
        bullet.gravity()
        bullet.move()
        bullet.collides(shooters,wall,bullist)
        bullet.out(surface,bullist)
        bullet.draw(surface)
    # wall
    wall.update()
    wall.draw(surface)
    pygame.draw.aaline(surface,(255,255,255),(0,400),(640,400))

def angle(A,B,BC,theta):

    C = [0,0]
    alpha = math.atan2(A[1]-B[1] , A[0] -B[0] ) - theta
    C[0] = int(round(B[0]  + BC * math.cos(alpha),0))
    C[1] = int(round(B[1]  + BC * math.sin(alpha),0))
    return C

class Shooter:
    def __init__(self,pos,size,color,xmax,xmin):
        self.pos = pos
        self.size = size
        self.rect = pygame.Rect(pos,size)
        self.health = 100
        self.color = color
        self.angle = 0
        self.speed = 0
        self.max = xmax
        self.min = xmin
        self.next_x = pos[0]

        self.color2 = []
        for i in color:
            i = i - 100
            if i < 0:
                i = 0
            self.color2.append(i)

    def draw(self,surface):
        global C
        pygame.draw.rect(surface,self.color,self.rect)

        c = angle(self.rect.midleft,self.rect.center,
                  20,radians(self.angle))
        if c != C and c != [95,392]:
            print c
            C = c
        pygame.draw.line(surface,self.color2,self.rect.center,c,3)


## the other funcs or classes not needed

# globals
turn = 1
C = []
# pygame
(width, height) = (640, 480)
screen = pygame.display.set_mode((width, height))
pygame.display.set_caption('Shooter')
clock = pygame.time.Clock()
# game actors
shooters = []
bullets = []
p1 = Shooter((400,400),(30,-15),(255,0,0),0,0)
p2 = Shooter((100,400),(30,-15),(0,255,0),0,0)
shooters.extend([p1, p2])
wall = Wall(100)

running = True
while running:
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            running = False
            pygame.quit()
            break
        else: event_handle(event,turn)
    if not running:
        break
    screen.fill((0,0,0))
    # Game draw logic + Game logic

    draw_all(bullets,shooters,wall,screen)
    pygame.display.flip()
    clock.tick(40)

我做错了什么?

2 个答案:

答案 0 :(得分:1)

您的代码确实有效,但存在一些问题。桶实际上是旋转的,但每次按键时只有很小的数量。

尝试将event_handle功能更改为:

def event_handle(turn):
    if turn == 1:
        c_s = p1
    elif turn == 2:
        c_s = p2
    else:
        return None
    key = pygame.key.get_pressed()
    # next_pos
    if key[pygame.K_q]:
        c_s.next_x -= 1
    if key[pygame.K_e]:
        c_s.next_x += 1
    # angle
    if key[pygame.K_w]:
        c_s.angle += radians(10)
    if key[pygame.K_s]:
        c_s.angle -= radians(10)
    # power (speed)
    if key[pygame.K_d]:
        c_s.speed += 0.1
    if key[pygame.K_a]:
        c_s.speed -= 0.1

由于此时你根本没有介入事件类型,我删除了event参数和if event.type == pygame.KEYDOWN:检查。这样,您可以按住按键,而不是被迫多次按键来旋转枪管。

我还将桶的值从radians(1)增加到radians(10)。否则,更改太小而无法显示(在合理的时间内)。

此外,您必须将主循环调整为

...
for event in pygame.event.get():
    if event.type == pygame.QUIT:
        running = False
        pygame.quit()
        break
if not running:
    break
event_handle(turn)
...

所以在主循环的每次迭代中调用event_handle

答案 1 :(得分:0)

如果您使用的是key.get_pressed,则根本不需要检测key_down事件。只需删除if event.type == pygame.KEYDOWN:语句,然后检查pygame.key.get_pressed返回的内容是否包含您的密钥。

Alos,您应该在每次循环运行时在代码中的某处调用pygame.event.pump。把它放在你的while循环中。

(和你的问题无关的提示:将一个主循环移动到一个函数中 - 将它挂在程序体中就像这样可怕了)