游戏用户输入在按回车后不会呈现

Question

我正试图在python中创建一个蠕虫克隆。我不是很有经验的游戏，并试图设置用户输入在屏幕上的小灰色框。但是，我使用的代码要么不正确地接受用户输入，要么不正确地呈现它。(我怀疑是后者。)我已经看过我多次引用的代码教程，检查了堆栈溢出帖子，但似乎没有一个有效。我希望在输入并按回键之后，单词会显示出来，因为为了代码简单起见，在键入时呈现的单词已经被移除。任何帮助都将不胜感激。

import random
import pygame

word_options = ("birth", "happy", "nancy")
word = random.choice(word_options).upper

WIDTH = 600
HEIGHT = 700
MARGIN = 10
T_MARGIN = 100
B_MARGIN = 100
LR_MARGIN = 100

GREY = (225, 227, 229)
GREY_FILLED = (120, 124, 126)
GREEN = (6,214,160)
YELLOW = (255, 209, 102)

INPUT = ""
GUESSES = ["berry"]
ALPHABET = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
UNGUESSED = ALPHABET
GAME_OVER = False

pygame.init()
pygame.font.init()
pygame.display.set_caption("Nancdle")

SQ_SIZE = (WIDTH-4*MARGIN-2*LR_MARGIN) // 5
FONT = pygame.font.SysFont("free sans bold", SQ_SIZE)
FONT_SMALL = pygame.font.SysFont("free sans bold", SQ_SIZE//2)

#screen
screen = pygame.display.set_mode((WIDTH, HEIGHT))
white = (255, 255, 255)
#animation loop
animating = True
while animating:

  screen.fill(white)
  letters =FONT_SMALL.render(UNGUESSED, False, GREY_FILLED)
  surface = letters.get_rect(center = (WIDTH//2, T_MARGIN//2))
  screen.blit(letters, surface)
  
  y = T_MARGIN
  for i in range(6):
    x = LR_MARGIN
    for j in range(5):
      square = pygame.Rect(x, y, SQ_SIZE, SQ_SIZE)
      pygame.draw.rect(screen, GREY, square, 2)
      #to add past letters/words
      if i < len(GUESSES):
        color = GREY_FILLED
        pygame.draw.rect(screen, color, square)
        letter = FONT.render(GUESSES[i][j], False, (255,255,255))
        surface = letter.get_rect(center = (x+SQ_SIZE//2, y+SQ_SIZE//2))
        screen.blit(letter, surface)
      x += SQ_SIZE + MARGIN
    y += SQ_SIZE + MARGIN
  
    
  #to update screen
  pygame.display.flip()
  for event in pygame.event.get():
    #so you can close window
    if event.type == pygame.QUIT:
      animating = False
    
    elif event.type == pygame.KEYDOWN:
      if event.key == pygame.K_RETURN:
        if len(INPUT) == 5:
          GUESSES.append(INPUT)
          GAME_OVER = True if INPUT == ANSWER else false
          INPUT = ""
      
    
        elif len(INPUT) < 5 and not GAME_OVER:
          INPUT = INPUT + event.unicode.upper()

如果这个问题措辞不当或者问错了，我很乐意删除这篇文章。

Answer 1

您的代码只处理K_Return的事件，而不是任何其他键。这是由于将elif len(INPUT) < 5 and not GAME_OVER:缩进过远造成的。

如果按以下方式更改事件处理程序代码，则应该可以工作：

elif event.type == pygame.KEYDOWN:
  if event.key == pygame.K_RETURN:
    if len(INPUT) == 5:
      GUESSES.append(INPUT)
      GAME_OVER = True if INPUT == word else False
      INPUT = ""

  elif len(INPUT) < 5 and not GAME_OVER:
    INPUT = INPUT + event.unicode.upper()

注:我还必须用ANSWER取代word，并将false资本化。

您使用两个空格进行缩进，如果使用了四个，则可能更容易识别范围不匹配。

使用event.unicode相当健壮，但用户仍然可以添加数字和空格。如果你想遍历所有的字母键，但不想要像if event.key in [pygame.K_a, pygame.K_b, ...这样可怕的东西，你可以使用range。

elif event.type == pygame.KEYDOWN:
    if event.key == pygame.K_RETURN:
        if len(INPUT) == 5:
            GUESSES.append(INPUT)
            GAME_OVER = True if INPUT == ANSWER else False
            INPUT = ""
    elif event.key in range(pygame.K_a, pygame.K_z + 1):
        # any other letter keys pressed
        if len(INPUT) < 5 and not GAME_OVER:
            INPUT = INPUT + event.unicode.upper()
    elif event.key in [pygame.K_BACKSPACE, pygame.K_DELETE]:
        INPUT = INPUT[:-1]  # don't need to check length