在pygame中分配密钥的便捷方法

Question

所以，我一直在构建这个琐事游戏，我想把键盘上的字母键分配给pygame。简单地说，我希望用户在选择答案时按a、b或c，然后pygame会拿起答案并告诉他们答案是对是错。请让我知道，因为我是一个初学者在建立基于pygame平台的游戏。

下面是我的代码片段：

    # Program: Import Library, Pygame, for initialization of this program
import pygame
import time
# Initialize the game engine
pygame.init()

# Define Colours

BLACK    = (   0,   0,   0)
WHITE    = ( 255, 255, 255)
GREEN    = (   0, 255,   0)
RED      = ( 255,   0,   0)
BLUE     = (   0,   0, 255)

display_width = 1080
display_height = 720
size = (display_width, display_height)
screen = pygame.display.set_mode(size)
pygame.display.set_caption("MiniConomy Trivia By Devang SAHANI")

# Button Program

class Button:
    def __init__(self, size, text, pos, bgColor=(0, 255, 0), textColor=(0, 0, 0)):
        self.pos  = pos
        self.size = size
        self.text = text
        self.font = pygame.font.Font(pygame.font.get_default_font(), size[1])
        self.textSurf = self.font.render(f"{text}", True, textColor)
        self.button = pygame.Surface((size[0], size[1])).convert()
        self.button.fill(bgColor)

    def render(self, window):
        window.blit(self.button, (self.pos[0], self.pos[1]))
        window.blit(self.textSurf, (self.pos[0]+1, self.pos[1]+5))

    def clicked(self, events):
        mousePos = pygame.mouse.get_pos()#  get the mouse position
        for event in events:
            if self.button.get_rect(topleft=self.pos).collidepoint(mousePos[0], mousePos[1]):
                if event.type == pygame.MOUSEBUTTONDOWN:
                    return True
        return False

# Setting a Title Screen
def text_objects(text, font):
    textSurface = font.render(text, True, BLACK)
    return textSurface, textSurface.get_rect()
largeText = pygame.font.Font('freesansbold.ttf', 90)

# Setting background sound
def background_sound():
    pygame.mixer.Sound.play(start_sound)
 
# Creating a Title Screen
TextSurf, TextRect = text_objects("MiniConomy", largeText)
TextRect.center = (540,150)

# Button Control

button = Button([280,50], "Let's Begin", [380,302])
button2 = Button([190, 50], "About", [380, 402])
button3 = Button([215, 50], "Settings", [380, 502])

#Loop until the user clicks the close button
done = False
# Used to manage how fast the screen updates
clock = pygame.time.Clock()

# Menu Settings

background_image = pygame.image.load("Miniconomy.PNG").convert()
about_image = pygame.image.load("abouthtp.PNG").convert()
start_sound = pygame.mixer.Sound("start.ogg")

# Question Image Sources

img_q1 = pygame.image.load("question_1.PNG").convert()
img_q2 = pygame.image.load("question_2.PNG").convert()
img_q3 = pygame.image.load("question_3.PNG").convert()
img_q4 = pygame.image.load("question_4.PNG").convert()
img_q5 = pygame.image.load("question_5.PNG").convert()
img_q6 = pygame.image.load("question_6.PNG").convert()
img_c = pygame.image.load("correct.PNG").convert()
img_ic = pygame.image.load("incorrect.PNG").convert()
img_exi = pygame.image.load("exitscreen.PNG").convert()

# -------- Main Program Loop -----------
screen.blit(background_image, (0, 0))
keep_buttons = True
while not done:
    events = pygame.event.get()
    for event in events: # User did something
        if event.type == pygame.QUIT: # If user clicked close
            done = True # Flag that we are done so we exit this loop

    # --- Game logic should go here
    # --- Drawing code should go here
    
 
    # Set the screen background
    if keep_buttons:
        screen.blit(TextSurf, TextRect)
    
    # Button 1 Control
    if keep_buttons:
        button.render(screen)
    if button.clicked(events):
        pygame.mixer.music.stop()
        screen.fill((0,0,0))
        time.sleep(1)
        screen.blit(img_q1, (0,0)) #This is where I want the answer options to be
        keep_buttons = False

    if keep_buttons:
        button2.render(screen)
    if button2.clicked(events):
        pygame.mixer.music.stop()
        screen.fill((0,0,0))
        screen.blit(about_image, (0,0))
        keep_buttons = False

    if keep_buttons:
        button3.render(screen)
    if button3.clicked(events):
        pygame.mixer.music.stop()
        print("Game logic goes here")
        pass

    # --- Go ahead and update the screen with what we've drawn.
    pygame.display.flip()
    # --- Limit to 60 frames per second
    clock.tick(60)

pygame.quit()
quit()

Answer 1

基本上，您需要做的是使用for event in pygame.events.get (以下称为更新节)并使用key属性编写。如果你这样写：

if event.type == pygame.KEYUP: # checking if the event has a key that was pressed - it's key up 
# since you don't want someone to accidentally skip through questions through long-pressing it

    if event.key == pygame.K_a: # checking if the key is 'a'

        # code to do if the 'a' key is pressed

使用它，您可以使每个键都可以做任何事情(您可以随时查看PyGame文档)。然而，我建议您使用函数或类来组织您的项目，因为它们使一些问题变得非常容易，便于以后使用(例如，您可以创建一个类，让它显示每个问题和答案，而不是硬编码)。

无论如何，祝你的项目好运。如果你需要更多的帮助，尽管说。