我是python/ 的新手pygame，我无法弄清楚。每当我按住一个键时，它就不会循环播放KEYDOWN. 另外，如果我按住键盘上的键并同时移动鼠标，它似乎会连续移动。有人能告诉我我做错了什么吗？import pygameimport randompygame.init()#Colorswhite = 255, 255, 255black = 0, 0, 0back_color = 48, 255, 124light_color = 34, 155, 78#Display W/Hdisplay_width = 800display_height = 600#X/Yx_axis = 400y_axis = 580Block_size = 20x_int = 0y_int = 0ON = TrueWindow = pygame.display.set_mode((display_width, display_height))pygame.display.set_caption('Game')#On Loopwhile ON == True: #Screen Event for Screen in pygame.event.get(): #Quit Screen if Screen.type == pygame.QUIT: pygame.quit() exit() #Quit Full Screen if Screen.type == pygame.KEYDOWN: if Screen.key == pygame.K_q: pygame.quit() exit() #Full Screen !!!!!!!! EDIT THIS !!!!!!!! if Screen.type == pygame.KEYDOWN: if Screen.key == pygame.K_1: pygame.display.set_mode((display_width, display_height),pygame.FULLSCREEN) if Screen.key == pygame.K_2: pygame.display.set_mode((display_width, display_height)) #Player Movement K DOWN if Screen.type == pygame.KEYDOWN: if Screen.key == pygame.K_d: x_int = 20 if Screen.key == pygame.K_a: x_int = -20 #Player Movement K UP if Screen.type == pygame.KEYUP: if Screen.key == pygame.K_d or Screen.key == pygame.K_a: x_int = 0 x_axis += x_int Window.fill((back_color)) Player = pygame.draw.rect(Window, light_color, [x_axis, y_axis, Block_size, Block_size]) pygame.display.update()quit()
2 回答
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TA贡献1789条经验 获得超10个赞
我已经改进了你的代码。您已将屏幕更新（绘制屏幕）部分放在事件循环中，而它应该在while循环中。我有模式的代码有点复杂，但按预期工作。为什么复杂？按住该键时，事件列表为空（您可以打印事件）。我也做了块不要走出屏幕。块的速度很高，所以我将其降低到10.
import pygame
import random
pygame.init()
#Colors
white = 255, 255, 255
black = 0, 0, 0
back_color = 48, 255, 124
light_color = 34, 155, 78
#Display W/H
display_width = 800
display_height = 600
#X/Y
x_axis = 400
y_axis = 580
global x_int
Block_size = 20
x_int = 0
y_int = 0
ON = True
Window = pygame.display.set_mode((display_width, display_height))
pygame.display.set_caption('Game')
global topass_a,topass_d
x_int,topass_a,topass_d = 0,0,0
#On Loop
while ON:
#Screen Event
events = pygame.event.get()
def on_a_press():
global topass_a,x_int
x_int = -10
topass_a = 1
def on_d_press():
global topass_d,x_int
x_int = 10
topass_d = 1
if len(events) == 0:
if topass_a == 1:on_a_press()
if topass_d == 1:on_d_press()
for Screen in events:
if Screen.type == pygame.QUIT:
pygame.quit()
exit()
if Screen.type == pygame.KEYDOWN:
if Screen.key == pygame.K_q:
pygame.quit()
exit()
if Screen.key == pygame.K_1:
pygame.display.set_mode((display_width, display_height),pygame.FULLSCREEN)
if Screen.key == pygame.K_2:
pygame.display.set_mode((display_width, display_height))
if Screen.key == pygame.K_d or topass_d == 1:
on_d_press()
if Screen.key == pygame.K_a or topass_a == 1:
on_a_press()
if Screen.type == pygame.KEYUP:
if Screen.key == pygame.K_d or Screen.key == pygame.K_a:
x_int = 0
topass_a = 0
topass_d = 0
x_axis += x_int
x_int = 0
if x_axis < 0:x_axis=0
elif x_axis >= display_width-Block_size:x_axis = display_width-Block_size
Window.fill((back_color))
Player = pygame.draw.rect(Window, light_color, [x_axis, y_axis, Block_size, Block_size])
pygame.display.update()
您可以根据需要进一步改进代码。
编辑：
为什么复杂？容易的事情是第一位的。我已经意识到没有必要跟踪密钥。pygame.key.get_pressed()返回按下的键。这是一个更小、更好和改进的代码。我还实现了w和s(y_axis) 键。
import pygame
import random
pygame.init()
#Colors
white = 255, 255, 255
black = 0, 0, 0
back_color = 48, 255, 124
light_color = 34, 155, 78
#Display W/H
display_width = 800
display_height = 600
#X/Y
x_axis = 400
y_axis = 580
Block_size = 20
x_int = 0
y_int = 0
ON = True
Window = pygame.display.set_mode((display_width, display_height))
pygame.display.set_caption('Game')
while ON:
events = pygame.event.get()
for Screen in events:
if Screen.type == pygame.QUIT:
pygame.quit()
exit()
if Screen.type == pygame.KEYDOWN:
if Screen.key == pygame.K_q:
pygame.quit()
exit()
if Screen.key == pygame.K_1:
pygame.display.set_mode((display_width, display_height),pygame.FULLSCREEN)
if Screen.key == pygame.K_2:
pygame.display.set_mode((display_width, display_height))
keys = pygame.key.get_pressed()
if keys[pygame.K_a]:
x_int = -10
if keys[pygame.K_d]:
x_int = 10
# keys controlling y axis, you can remove these lines
if keys[pygame.K_w]:
y_int = -10
if keys[pygame.K_s]:
y_int = 10
#x_axis......
x_axis += x_int
x_int = 0
if x_axis < 0:x_axis=0
elif x_axis >= display_width-Block_size:x_axis = display_width-Block_size
#y axis
y_axis += y_int
y_int = 0
if y_axis < 0:y_axis=0
elif y_axis >= display_height-Block_size:y_axis = display_height-Block_size
#updaing screen
Window.fill((back_color))
Player = pygame.draw.rect(Window, light_color, [x_axis, y_axis, Block_size, Block_size])
pygame.display.update()
慕后森
TA贡献1802条经验 获得超5个赞
您仅pygame.KEYDOWN在第一次按下该键时接收- 而不是在按住时接收。简单的解决方案是只在键按下时绘制（即 when x_int != 0）
#On Loop
while ON == True:
#Screen Event
for Screen in pygame.event.get():
# <Removed not relevant code for brevity>
#Player Movement K DOWN
if Screen.type == pygame.KEYDOWN:
if Screen.key == pygame.K_d:
x_int = 20
if Screen.key == pygame.K_a:
x_int = -20
#Player Movement K UP
if Screen.type == pygame.KEYUP:
if Screen.key == pygame.K_d or Screen.key == pygame.K_a:
x_int = 0
# Render only happens if x_int is not zero
# (Need to add code to force render first time)
if x_int:
x_axis += x_int
Window.fill((back_color))
Player = pygame.draw.rect(Window, light_color, [x_axis, y_axis, Block_size, Block_size])
pygame.display.update()
随着程序的增长和变得越来越复杂，您需要更好的逻辑来确定何时渲染，但这会让您开始。
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