2 回答

TA貢獻1789條經驗 獲得超10個贊
我已經改進了你的代碼。您已將屏幕更新(繪制屏幕)部分放在事件循環中,而它應該在while循環中。我有模式的代碼有點復雜,但按預期工作。為什么復雜?按住該鍵時,事件列表為空(您可以打印事件)。我也做了塊不要走出屏幕。塊的速度很高,所以我將其降低到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()
您可以根據需要進一步改進代碼。
編輯:
為什么復雜?容易的事情是第一位的。我已經意識到沒有必要跟蹤密鑰。pygame.key.get_pressed()返回按下的鍵。這是一個更小、更好和改進的代碼。我還實現了w和s(y_axis) 鍵。
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()

TA貢獻1802條經驗 獲得超5個贊
您僅pygame.KEYDOWN在第一次按下該鍵時接收- 而不是在按住時接收。簡單的解決方案是只在鍵按下時繪制(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()
隨著程序的增長和變得越來越復雜,您需要更好的邏輯來確定何時渲染,但這會讓您開始。
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