如何在 pygame 中连续循环 KEYDOWN？
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 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')
#On Loop
while 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()
Cats萌萌2回答
-
至尊宝的传说
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 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 = 580global x_intBlock_size = 20x_int = 0y_int = 0ON = TrueWindow = pygame.display.set_mode((display_width, display_height))pygame.display.set_caption('Game')global topass_a,topass_dx_int,topass_a,topass_d = 0,0,0#On Loopwhile 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 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')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
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 Loopwhile 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。
随时随地看视频慕课网APP
相关分类
Python