首页 > 解决方案 > 如何检测我的两个图像是否在 Pygame 中接触?

问题描述

我正在制作一款 ASL 手势掉落的游戏,您需要将它们接在桶中。当标志接触桶时,我希望它打印出他们正在接触。如果他们不是,我希望它打印出他们没有接触。如何检测图像是否接触?我已经有一些矩形碰撞代码,但它只是不断地打印出它们正在触摸其中一个标志,即使它们不是。提前感谢您能给我的任何帮助!PS 很抱歉输入了我所有的代码,但我不确定需要查看什么来帮助解决我的问题。

我查看了许多涉及图像碰撞的问题和答案,但我似乎无法让它在我的代码中工作。我“突出显示”了我认为导致问题的代码。

import random
import pygame
pygame.init()
pygame.mixer.init()
#define functions
def newPlacement():
    screen.blit(choiceASL, (x,y))
    screen.blit(choiceASL2, (x2,y2))
    screen.blit(choiceASL3, (x3,y3))
def bucketStuff():
    screen.blit(bucketPic, (bucketX, bucketY))
'''
def ifCollided():
    if bucketRect.colliderect(choiceASLRect):
        print("You collided with choice1!")
    elif bucketRect.colliderect(choiceASL2Rect):
        print("You collided with choice2!")
    elif bucketRect.colliderect(choiceASL3Rect):
        print("You collided with choice3!")
    else:
        print("Not touching.")
'''

#define stuff below: var, funct, def, etc.
aslABCSList = ("AslA.png", "AslB.png", "AslC.png", "AslD.png")
randHandsign = random.choice(aslABCSList)
randHandsign2 = random.choice(aslABCSList)
randHandsign3 = random.choice(aslABCSList)
choiceASL = pygame.image.load(randHandsign)
choiceASLRect = choiceASL.get_rect()
choiceASL2 = pygame.image.load(randHandsign2)
choiceASL2Rect = choiceASL2.get_rect()
choiceASL3 = pygame.image.load(randHandsign3)
choiceASL3Rect = choiceASL3.get_rect()
bucketPic = pygame.image.load("bucketPic.png")
bucketRect = bucketPic.get_rect()

screen = pygame.display.set_mode((700, 600))
Black = [0, 0, 0]
White = [255, 255, 255]
y = 0
y2 = 0
y3 = 0
x = int(random.randrange(1,650,25))
x2 = int(random.randrange(1,650,25))
x3 = int(random.randrange(1,650,50))
bucketX = 300
bucketY = 540

#to make the main work
done = False
#allows fps rate
clock = pygame.time.Clock()

#basically main, put anything that you want to run consistently until the user exits below.
while not done:
        for event in pygame.event.get():
                if event.type == pygame.QUIT:
                        done = True
        #vvv Put all main code here! vvv
        screen.fill(White)
        bucketStuff()
        newPlacement()
        pressed = pygame.key.get_pressed()
        if pressed[pygame.K_LEFT]: bucketX -= 3
        if pressed[pygame.K_RIGHT]: bucketX += 3
        ifCollided()

        y+=1
        y2+=2
        y3+=3



        #below is basically a refresh, so that everything is updated. Put code before this!
        pygame.display.flip()
        #60 fps
        clock.tick(50)

标签: pythonimagepygamecollision-detection

解决方案


对象的返回.get_rect()一个pygame.Surface具有表面大小的矩形,但位于 (0, 0) 位置。
您必须设置图像的位置 ( pygame.Rect):

例如:

choiceASL = pygame.image.load(randHandsign)
choiceASLRect = choiceASL.get_rect()

# [...]

bucketPic = pygame.image.load("bucketPic.png")
bucketRect = bucketPic.get_rect()

# [...]

while not done:

    # [...]

    if pressed[pygame.K_LEFT]: bucketX -= 3
    if pressed[pygame.K_RIGHT]: bucketX += 3

    bucketRect.topleft = (bucketX, bucketY)
    choiceASLRect.topleft = (x, y)
    # [...]

    ifCollided()

一旦正确设置了矩形的位置,碰撞测试将按预期工作。


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