这个错误是什么意思:TypeError: argument 1 must be pygame.Surface, not pygame.Rect

问题描述

这是我的代码，它真的很简单:

Here is my code, it is really simple:

bif="C:\\Users\\Andrew\\Pictures\\pygame pictures\\Background(black big).png"
import pygame, sys
from pygame.locals import *

pygame.init()

screen=pygame.display.set_mode((1000,1000),0,32)
background=pygame.image.load(bif).convert()
line=pygame.draw.line(background, (255,255,255), (30,31), (20,31))
while True:

    mousex, mousey = pygame.mouse.get_pos()
    for event in pygame.event.get():
        if event.type == QUIT:
            pygame.quit()
            sys.exit()
    screen.blit(background, (0,0))
    screen.blit(line,(mousex,mousey))
    pygame.display.update()

这样做的目的是让线条在屏幕上跟随我的鼠标.但是，每当我尝试运行该程序时，都会出现错误:

The goal of this is to have the line follow my mouse on the screen. However, whenever I try to run the program I get the error :

TypeError: 参数 1 必须是 pygame.Surface，而不是 pygame.Rect

TypeError: argument 1 must be pygame.Surface, not pygame.Rect

这是什么意思，我做错了什么?

What does this mean, and what am I doing wrong?

推荐答案

该错误意味着它期望的是表面而不是矩形.这是违规行:

The error means that it is expecting a surface not a rectangle. This is the offending line:

screen.blit(line,(mousex,mousey))

问题在于您将 draw.line() 的结果分配给 line.

The problem is that you are assigning the result of draw.line() to line.

line=pygame.draw.line(background, (255,255,255), (30,31), (20,31))

来自 pygame 文档:

From the pygame docs:

所有绘图函数都尊重 Surface 的裁剪区域，并且将被限制在该区域.函数返回一个矩形表示改变像素的边界区域.

所以您要为线分配一个矩形.不确定您到底要做什么，但您可以通过以下方式绘制跟随鼠标的线:

So you are assigning a rectangle to line. Not sure exactly what you want to do, but you can draw the line following the mouse in the following way:

pygame.draw.line(background, (255,255,255), (x,y), (x,y+10))

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