没有错误消息，但这不会输出任何内容

编程问答 2022-05-02

问题描述

不要问我在这里想做什么，但是如果我只绘制一个没有类的矩形，请帮忙，它工作正常，但不输出任何内容，甚至没有错误消息。那我应该不上课吗？我对这种东西很糟糕，它告诉我我需要写更多的字，而且我的帖子主要是代码，所以忽略这些东西 rn

import pygame
import os
pygame.init()
 
FPS=60

SCREEN = pygame.display.set_mode((400,500))
pygame.display.set_caption('test')

  
x=50
y=450
vel = 3
width = 20
height = 20
walk = pygame.draw.rect(SCREEN,(0,0),(x,y,width,height))
class player(object):
    def __init__(self,x,height):
        self.x = x
        self.y = y
        self.width = width
        self.height = height
        self.vel = 3
    def draw(self,SCREEN):
        SCREEN.blit (walk,(self.x,self.y))
        
    def rdw():
        SCREEN.blit((0,0))
        player.draw(SCREEN)
        pygame.display.update()

player = player(x,height)
#main loop
running = True
clock = pygame.time.Clock()


while running:
    clock.tick(FPS)

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            running = False

        
        keys = pygame.key.get_pressed()

    if keys[pygame.K_a] and player.x > 30 - player.width - player.vel:
                x-=player.vel
    if keys[pygame.K_d] and player.x < 400 - player.width - player.vel:
                x+=player.vel



    SCREEN.fill((255,255,255))
pygame.display.update()

pygame.quit()

解决方法

pygame.display.update() 需要缩进到 while running 循环内。还需要在填满屏幕后调用 player.draw 方法。

此外，SCREEN.blit((0,0)) 会导致程序崩溃，因为您没有为它提供任何要 blit 的东西。

,

创建玩家的 pygame.Surface 对象大小，并使用玩家的颜色 fill 它。这是您需要在 Player.draw 中 blit 的对象：

class Player(object):
    def __init__(self,x,y,width,height):
        # [...]
        
        self.image = pygame.Surface((width,height))
        self.image.fill((0,0))
    
    def draw(self,SCREEN):
        SCREEN.blit(self.image,(self.x,self.y))

player = Player(x,height)

移动播放器时，需要更改player（player.x，player.y）而不是x和y的坐标： 

keys = pygame.key.get_pressed()
if keys[pygame.K_a] and player.x > 30 - player.width - player.vel:
    player.x -= player.vel
if keys[pygame.K_d] and player.x < 400 - player.width - player.vel:
    player.x += player.vel

您必须在清除屏幕后和更新屏幕之前绘制播放器：

SCREEN.fill((255,255,255))
player.draw(SCREEN)
pygame.display.update()

完整示例（关心 Indentation）：

import pygame
import os
pygame.init()
 
FPS=60

SCREEN = pygame.display.set_mode((400,500))
pygame.display.set_caption('test')

x=50
y=450
vel = 3
width = 20
height = 20

class Player(object):
    def __init__(self,height):
        self.x = x
        self.y = y
        self.width = width
        self.height = height
        self.vel = 3
        self.image = pygame.Surface((width,0))
    def draw(self,height)

#main loop
running = True
clock = pygame.time.Clock()
while running:
    clock.tick(FPS)

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            running = False
 
    keys = pygame.key.get_pressed()
    if keys[pygame.K_a] and player.x > 30 - player.width - player.vel:
        player.x -= player.vel
    if keys[pygame.K_d] and player.x < 400 - player.width - player.vel:
        player.x += player.vel

    SCREEN.fill((255,255))
    player.draw(SCREEN)
    pygame.display.update()

pygame.quit()

典型的 PyGame 应用程序循环必须：

classclassclasspygamepython