我编写了以下代码，创建了一个简单的游戏，当你点击键盘上的箭头时，一个盒子在游戏中移动一个单元。
我试图这样做，以便如果我按下任何箭头按钮，该框将继续向该方向移动，直到推动另一个箭头。因此，如果我按下右箭头而不是拍摄+50像素，它将在屏幕上连续移动，直到点击一个不同的箭头，然后就会这样了
import pygame #importing the pygame library
# some initializations
pygame.init() # this line initializes pygame
window = pygame.display.set_mode( (800,600) ) # Create a window with width=800 and height=600
pygame.display.set_caption( 'Rectangle move' ) # Change the window's name we create to "Rectangle move"
clock = pygame.time.Clock() # Clocks are used to track and control the frame-rate of a game (how fast and how slow the pace of the game)
# This line creates and initializes a clock.
# color definitions, using RBG color model.
black = (0,0,0)
white = (255,255,255)
# initial center position for the square (bob)
x,y = 0,0
lastKey=0
game_loop=True
while game_loop:
for event in pygame.event.get(): # loop through all events
if event.type == pygame.QUIT:
game_loop = False # change the game_loop boolean to False to quit.
if event.type == pygame.KEYDOWN:
lastKey = event.key
#check last entered key
#lastKey equals "LEFT", "RIGHT", "UP", "DOWN" --> do the required stuff!
#set x coordinate minus 50 if left was pressed
if lastKey == pygame.K_LEFT:
x -= 50
if lastKey == pygame.K_RIGHT:
x += 50
if lastKey == pygame.K_UP:
y += 50
if lastKey == pygame.K_DOWN:
y -= 50
if event.key == pygame.K_LEFT:
x -= 50
if event.key == pygame.K_RIGHT:
x += 50
if event.key == pygame.K_UP:
y += 50
if event.key == pygame.K_DOWN:
y -= 50
# draw and update screen
window.fill( black ) # fill the screen with black overwriting even bob.
pygame.draw.rect( window, white, (x, y, 50, 50) ) # draw bob on the screen with new coordinates after its movement.
# the parameters are as follows: window: is the window object you want to draw on. white: the object color used to fill the rectangle
# (x,y,50,50) x is the x position of the left side of the rectangle. y is the y position of the upper side of the rectangle.
# In other words (x,y) is the coordinate of the top left point of the rectangle.
# 50 is the width, and 50 is the height
pygame.display.update() #updates the screen with the new drawing of the rectangle.
#fps stuff:
clock.tick(10) # this controls the speed of the game. low values makes the game slower, and large values makes the game faster.
pygame.quit()
任何帮助都会非常感激。
答案 0 :(得分：2)
您想要在按某个键时更改应用程序的状态。所以你需要一个变量来跟踪那个状态（状态是：框应该移动什么方向？）。
这是一个完整的，最小的例子，可以满足您的需求。请注意评论。
import pygame, sys
pygame.init()
screen = pygame.display.set_mode((640, 480))
screen_r = screen.get_rect()
clock = pygame.time.Clock()
rect = pygame.rect.Rect(0, 0, 50, 50)
# let's start at the center of the screen
rect.center = screen_r.center
# a dict to map keys to a direction
movement = {pygame.K_UP: ( 0, -1),
pygame.K_DOWN: ( 0, 1),
pygame.K_LEFT: (-1, 0),
pygame.K_RIGHT: ( 1, 0)}
move = (0, 0)
# a simple helper function to apply some "speed" to your movement
def mul10(x):
return x * 10
while True:
for e in pygame.event.get():
if e.type == pygame.QUIT:
sys.exit()
# try getting a direction from our dict
# if the key is not found, we don't change 'move'
if e.type == pygame.KEYDOWN:
move = movement.get(e.key, move)
# move the rect by using the 'move_ip' function
# but first, we multiply each value in 'move' with 10
rect.move_ip(map(mul10, move))
# ensure that 'rect' is always inside the screen
rect.clamp_ip(screen_r)
screen.fill(pygame.color.Color('Black'))
pygame.draw.rect(screen, pygame.color.Color('White'), rect)
pygame.display.update()
clock.tick(60)
我使用Rect而不是跟踪两个坐标x和y，因为这样可以使用move_ip和clamp_ip函数轻松移动屏幕内的矩形。
答案 1 :(得分：1)
尝试将输入的密钥保存到变量中，并在Event-Loop之后进行检查。 像这样：
#...
lastKey = None
while game_loop:
for event in pygame.event.get(): # loop through all events
if event.type == pygame.QUIT:
game_loop = False # change the game_loop boolean to False to quit.
if event.type == pygame.KEYDOWN:
lastKey = event.key
#check last entered key
#lastKey equals "LEFT", "RIGHT", "UP", "DOWN" --> do the required stuff!
#set x coordinate minus 50 if left was pressed
if lastKey == pygame.K_LEFT
x -= 50
#<add the other statements here>
#(...)
我建议不要使用那么多if语句。一段时间后，它可能会有点混乱。 请检查以下问题以保持代码简短：
答案 2 :(得分：0)
以下是两个版本，第一个演示了如何利用事件循环来获得连续的移动（类似于Sloth的解决方案，但对于不熟悉字典的初学者来说有点简单），第二个一个展示如何使用pygame.key.get_pressed()实现此目的。
解决方案1：检查事件循环中按下了哪个键，并将x和y速度更改为所需的值。然后将速度添加到while循环中的rect.x和rect.y位置。
我实际上建议使用vectors而不是velocity_x和velocity_y变量，而另一个变量用于精灵的实际位置。 pygame.Rect不能将浮点数作为坐标，因此位置的向量或单独变量会更准确。
import pygame as pg
def main():
screen = pg.display.set_mode((640, 480))
clock = pg.time.Clock()
rect = pg.Rect(100, 200, 40, 60)
velocity_x = 0
velocity_y = 0
done = False
while not done:
for event in pg.event.get():
if event.type == pg.QUIT:
done = True
elif event.type == pg.KEYDOWN:
if event.key == pg.K_d:
velocity_x = 4
elif event.key == pg.K_a:
velocity_x = -4
elif event.type == pg.KEYUP:
if event.key == pg.K_d and velocity_x > 0:
velocity_x = 0
elif event.key == pg.K_a and velocity_x < 0:
velocity_x = 0
rect.x += velocity_x
rect.y += velocity_y
screen.fill((40, 40, 40))
pg.draw.rect(screen, (150, 200, 20), rect)
pg.display.flip()
clock.tick(30)
if __name__ == '__main__':
pg.init()
main()
pg.quit()
解决方案2：致电pygame.key.get_pressed以检查当前正在按下哪个键。检查是否按住左，右，上或下键，然后调整每帧精灵的位置。
pygame.key.get_pressed的缺点是您无法知道按键的顺序，但代码看起来有点简单。
import pygame as pg
def main():
screen = pg.display.set_mode((640, 480))
clock = pg.time.Clock()
rect = pg.Rect(100, 200, 40, 60)
velocity = (0, 0)
done = False
while not done:
for event in pg.event.get():
if event.type == pg.QUIT:
done = True
keys = pg.key.get_pressed()
if keys[pg.K_d]:
rect.x += 4
if keys[pg.K_a]:
rect.x -= 4
screen.fill((40, 40, 40))
pg.draw.rect(screen, (150, 200, 20), rect)
pg.display.flip()
clock.tick(30)
if __name__ == '__main__':
pg.init()
main()
pg.quit()