Python Pygame:玩家获胜';按键按下时,请勿移动
从pygame开始,当你按下箭头键时,试着让一个简单的点在屏幕上移动。目前,它只在你按下键时移动,但你必须反复按下它Python Pygame:玩家获胜';按键按下时,请勿移动,python,pygame,Python,Pygame,从pygame开始,当你按下箭头键时,试着让一个简单的点在屏幕上移动。目前,它只在你按下键时移动,但你必须反复按下它 import random import pygame import keyboard import time from pygame.locals import * class Player: def __init__(self): self.x = 150 self.y = 150 pygame.init() screen = py
import random
import pygame
import keyboard
import time
from pygame.locals import *
class Player:
def __init__(self):
self.x = 150
self.y = 150
pygame.init()
screen = pygame.display.set_mode((400, 300))
pygame.display.set_caption("Smile, you're beautiful!")
player = Player()
while True:
pygame.time.Clock().tick(60)
for event in pygame.event.get():
if event.type == QUIT:
exit()
if event.type == pygame.KEYDOWN:
if event.key == pygame.K_RIGHT:
player.x += 5
if event.key == pygame.K_DOWN:
player.y += 5
if event.key == pygame.K_LEFT:
player.x -= 5
if event.key == pygame.K_UP:
player.y -= 5
pygame.event.pump()
pygame.display.flip()
pygame.display.update()
screen.fill((0,0,0))
pygame.draw.circle(screen, (180, 180, 180), (player.x, player.y), 5)
此外,如果您能提供有关我当前代码的任何提示,我将不胜感激,这些提示可能会得到改进或更改,以提高效率。您需要使用按键关闭事件,而不是按键关闭事件。通过这种方式,您可以知道当前在每个滴答声上按下的键
while True:
pressed = pygame.key.get_pressed()
if pressed[pygame.K_RIGHT]:
player.x += 5
if pressed[pygame.K_DOWN]:
player.y += 5
if pressed[pygame.K_LEFT]:
player.x -= 5
if pressed[pygame.K_UP]:
player.y -= 5
您需要使用而不是按键关闭事件。通过这种方式,您可以知道当前在每个滴答声上按下的键
while True:
pressed = pygame.key.get_pressed()
if pressed[pygame.K_RIGHT]:
player.x += 5
if pressed[pygame.K_DOWN]:
player.y += 5
if pressed[pygame.K_LEFT]:
player.x -= 5
if pressed[pygame.K_UP]:
player.y -= 5