如何在Pygame中获取键盘输入

Question

我正在尝试在程序上获取键盘输入，其中用户必须在屏幕上移动一个点。但是，当我运行该程序时，控制台没有出现。你能帮我解决这个错误吗？这是我的代码：

import pygame
import random
pygame.init()
x=0
y=0
xdir=5
ydir=5
r = random.randint(10,255)
g = random.randint(10,255)
b = random.randint(10,255)
screen = pygame.display.set_mode((500,300))
pygame.draw.circle(screen, (r, g, b), (x,y), 15, 0)
USI = pygame.key.get_pressed()
while True:
    pygame.draw.circle(screen, (r, g, b), (x,y), 15, 0)
    pygame.display.update()
    if USI[pygame.K_LEFT]:
        x -= 5
if USI[pygame.K_RIGHT]:
    x += 5
if USI[pygame.K_UP]:
    y -= 5
if USI[pygame.K_DOWN]:
    y += 5

Answer 1

deteck键的示例循环，用pygame按下：

# creating a running loop
while True:
       
    # creating a loop to check events that
    # are occuring
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            pygame.quit()
            sys.exit()
         
        # checking if keydown event happened or not
        if event.type == pygame.KEYDOWN:
               
            # checking if key "A" was pressed
            if event.key == pygame.K_a:
                print("Key A has been pressed")
               
            # checking if key "J" was pressed
            if event.key == pygame.K_j:
                print("Key J has been pressed")
               
            # checking if key "P" was pressed
            if event.key == pygame.K_p:
                print("Key P has been pressed")
             
            # checking if key "M" was pressed
            if event.key == pygame.K_m:
                print("Key M has been pressed")

Answer 2

import pygame

pygame.init()
window = pygame.display.set_mode((300, 300))
clock = pygame.time.Clock()

rect = pygame.Rect(0, 0, 20, 20)
rect.center = window.get_rect().center
vel = 5

run = True
while run:
    clock.tick(60)
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            run = False
        if event.type == pygame.KEYDOWN:
            print(pygame.key.name(event.key))

    keys = pygame.key.get_pressed()
    
    rect.x += (keys[pygame.K_RIGHT] - keys[pygame.K_LEFT]) * vel
    rect.y += (keys[pygame.K_DOWN] - keys[pygame.K_UP]) * vel
        
    rect.centerx = rect.centerx % window.get_width()
    rect.centery = rect.centery % window.get_height()

    window.fill(0)
    pygame.draw.rect(window, (255, 0, 0), rect)
    pygame.display.flip()

pygame.quit()
exit()

https://replit.com/@Rabbid76/PyGame-ContinouslyMovement#main.py< /a>

这应该可以帮助你弄清楚 pygame 中的移动。

Answer 3

您可以从pygame获得事件，然后注意键盘事件，而不是查看get_pressed（）返回的键那个框架）。
您的代码现在正在发生的事情是，如果您的游戏以30fps渲染，并且您将左箭头键持续半秒，则您将在15次更新位置。

events = pygame.event.get()
for event in events:
    if event.type == pygame.KEYDOWN:
        if event.key == pygame.K_LEFT:
            location -= 1
        if event.key == pygame.K_RIGHT:
            location += 1

为了支持钥匙的持续移动，您必须建立某种限制，要么基于强制的游戏循环帧速率，要么是由您只允许您移动的每一个如此多的tick的计数器环形。

move_ticker = 0
keys=pygame.key.get_pressed()
if keys[K_LEFT]:
    if move_ticker == 0:
        move_ticker = 10
        location -= 1
        if location == -1:
            location = 0
if keys[K_RIGHT]:
    if move_ticker == 0:   
        move_ticker = 10     
        location+=1
        if location == 5:
            location = 4

然后，在游戏循环中的某个地方，您会做这样的事情：

if move_ticker > 0:
    move_ticker -= 1

这只会让您每10帧移动一次（因此，如果您移动，则记录器将设置为10帧，并且在10帧后，它将允许您再次移动）