无法在 Pygame 中跨屏幕移动图像

Question

我是 Python 的初学者，我正在尝试想出一个函数来在 Pygame 的帮助下绘制一个在屏幕上水平和垂直移动的正方形。当我们单击键盘上的箭头时，正方形应该会移动。

我的问题：虽然正方形显示在屏幕上，但它根本不动，但 Python 没有显示任何类型的错误。

有人知道缺少什么吗？提前致谢！

这是我目前所写的：

import pygame

altura_tela = 600
largura_tela = 800
tela = pygame.display.set_mode((largura_tela, altura_tela))

vermelho = (255, 0, 0)

altura_quadrado = 100
largura_quadrado = 100

def desenhaQuadrado(quadrado):
    pygame.draw.rect(tela, vermelho, (350,250,largura_quadrado, altura_quadrado))

def posicaoquadrado(quadradoPos):
    tela.blit(quadrado, (quadradoPos[0],quadradoPos[1]))

def moveQuadrado(teclas, quadradoPos):
    if teclas[0] and quadradoPos[1] > 0:
        quadradoPos[1] -= 20
    elif teclas[2] and quadradoPos[1] < 420:
        quadradoPos[1] += 20
    if teclas[1] and quadradoPos[0] > 0:
        quadradoPos[0] -= 20
    elif teclas[3]and quadradoPos[0] < 570:
        quadradoPos[0] += 20
    return quadradoPos

def main():
    pygame.init()
    teclas = [False, False, False, False]
    quadradoPos = [350,250]

    pygame.display.set_caption('Movimentação do Quadrado')

    terminou = False
    while not terminou:

        for event in pygame.event.get():

            if event.type == pygame.QUIT:
                terminou = True

            if event.type == pygame.KEYDOWN:
                if event.key == pygame.K_UP:
                    teclas[0] = True
                elif event.key == pygame.K_DOWN:
                    teclas[1] = True
                elif event.key == pygame.K_LEFT:
                    teclas[2] = True
                elif event.key == pygame.K_RIGHT:
                    teclas[3] = True

            if event.type == pygame.KEYUP:
                if event.key == pygame.K_UP:
                    teclas[0] = False
                elif event.key == pygame.K_DOWN:
                    teclas[1] = False
                elif event.key == pygame.K_LEFT:
                    teclas[2] = False
                elif event.key == pygame.K_RIGHT:
                    teclas[3] = False

            quadradoPos = moveQuadrado(teclas, quadradoPos)
            desenhaQuadrado(quadradoPos)

            pygame.display.update()


    pygame.display.quit()

    pygame.quit()

if __name__ == '__main__':
    main()

Answer 1

在你的函数中：

def desenhaQuadrado(quadrado):
    pygame.draw.rect(tela, vermelho, (350,250,largura_quadrado, altura_quadrado))

您正在传递一个参数quadrado，但您从未使用它。此外，您的坐标永远不会更新350, 250。所以这个函数基本上一遍又一遍地绘制相同的矩形。

我修改了你的代码：

import pygame

clock = pygame.time.Clock()

altura_tela = 600
largura_tela = 800
tela = pygame.display.set_mode((largura_tela, altura_tela))

FPS = 30

vermelho = (255, 0, 0)

altura_quadrado = 100
largura_quadrado = 100


def redraw(quadradoPos):
    tela.fill(pygame.Color("white"))
    pygame.draw.rect(tela, vermelho, (quadradoPos[0], quadradoPos[1], largura_quadrado, altura_quadrado))
    pygame.display.update()


def main():
    pygame.init()
    quadradoPos = [300,250]
    pygame.display.set_caption('Movimentação do Quadrado')

    terminou = False
    while not terminou:

        clock.tick(FPS)

        for event in pygame.event.get():

            if event.type == pygame.QUIT:
                terminou = True

            if event.type == pygame.KEYDOWN:
                if event.key == pygame.K_UP and quadradoPos[1] > 0:
                    quadradoPos[1] -= 50
                elif event.key == pygame.K_DOWN and quadradoPos[1] <= altura_tela - altura_quadrado:
                    quadradoPos[1] += 50
                elif event.key == pygame.K_LEFT and quadradoPos[0] >= 0:
                    quadradoPos[0] -= 50
                elif event.key == pygame.K_RIGHT and quadradoPos[0] <= largura_tela - largura_quadrado:
                    quadradoPos[0] += 50

            redraw(quadradoPos)

    pygame.quit()

if __name__ == '__main__':
    main()

注意：

-不要忘记tela.fill(pygame.Color("white"))，否则你会得到屏幕上绘制的所有矩形的“痕迹”。

-我添加clock是为了控制帧率。

-您不能像在posicaoquadrado(quadradoPos) 中那样在屏幕上闪烁rect，这将导致TypeError。您只能对 Surface 进行 blit。

希望这能回答你的问题。

Answer 2

desenhaQuadrado 函数内部有一些静态值

def desenhaQuadrado(quadrado):
    pygame.draw.rect(tela, vermelho, (350,250,largura_quadrado, altura_quadrado))

把它改成下面的代码，它就会移动。

def desenhaQuadrado(quadrado):
    pygame.draw.rect(tela, vermelho, (quadrado[0],quadrado[1],largura_quadrado, altura_quadrado))

尽管您将看到其他一些问题需要解决。 我创建了一个基本上是您想要的脚本，如果您需要帮助，请检查它 https://github.com/eng-robsonsampaio/pygame_move_square.git

import pygame
from pygame.locals import *

UP = 0
DOWN = 1
LEFT = 2
RIGHT = 3
GAME_ON = True
SPEED = 20

screen = pygame.display.set_mode((400, 400))
clock = pygame.time.Clock()

square = pygame.Surface((50,50))
square.fill((255,0,0))
POS = [(200,200)]

def move_square(direction):
    if UP == direction:
        POS.append((POS[0][0], POS[0][1] - 20))
    elif direction == DOWN:
        POS.append((POS[0][0], POS[0][1] + 20))
    elif direction == LEFT:
        POS.append((POS[0][0] - 20, POS[0][1]))
    else:
        POS.append((POS[0][0] + 20, POS[0][1]))
    POS.pop(0)


while GAME_ON:

    for event in pygame.event.get():
        if event.type == QUIT:
            GAME_ON = False
        if event.type == KEYDOWN:
            if event.key==K_UP:
                print("UP")
                move_square(UP)                
            elif event.key==K_LEFT:
                print("LEFT")
                move_square(LEFT)               
            elif event.key==K_DOWN:
                print("DOWN")
                move_square(DOWN)                
            elif event.key==K_RIGHT:
                print("RIGHT")
                move_square(RIGHT)

    screen.fill((0,0,0))
    screen.blit(square, POS[0])
    pygame.display.update()

pygame.quit()

我还有一个中号蛇游戏教程： https://medium.com/@robsonsampaio90/snake-game-in-python-with-pygame-291f5206a35e

您可能会在此链接中找到一些帮助。