我正试图让我的绿色方块检测到与蓝色的碰撞 但是,我不知道如何构造if语句,以便它在接触时立即发生碰撞。
如果我这样做, rect1 的位置是> = 200,100 如果越过蓝色正方形,它也会检测到碰撞
这是我的代码:
import pygame, sys
import time
import random
pygame.init()
screen = pygame.display.set_mode((640,480)) #Display
running = True
randomList = ("Hello", "Hi", "Why", "Die", "Billy Nye")
#Colors
white = (255,255,255)
black = (0,0,0)
red = (255,0,0)
blue = (0,0,255)
green = (0,255,0)
#Time variables
time = pygame.time.Clock()
FPS = 60
#Movement Variables
lead_x = 300
lead_y = 200
lead_x1 = 200
lead_y1 = 100
x_change = 0
y_change = 0
position = (200,100)
#Running MAIN Loop
while running:
for event in pygame.event.get():
if event.type == pygame.QUIT:
running = False
screen.fill(white)
rect2 = pygame.draw.rect(screen, green, [lead_x,lead_y, 50, 50])
rect1 = pygame.draw.rect(screen, blue, [lead_x1,lead_y1, 50 ,50])
pygame.display.update()
if event.type == pygame.KEYDOWN:
if event.key == pygame.K_LEFT:
x_change = -10
if event.key == pygame.K_RIGHT:
x_change = 10
if event.key == pygame.K_UP:
y_change = -10
if event.key == pygame.K_DOWN:
y_change = 10
if event.type == pygame.KEYUP:
if event.key == pygame.K_LEFT:
x_change = 0
if event.key == pygame.K_RIGHT:
x_change = 0
if event.key == pygame.K_UP:
y_change = 0
if event.key == pygame.K_DOWN:
y_change = 0
if lead_x == lead_x1 and lead_y == lead_y1:
print (random.choice(randomList))
lead_x += x_change
lead_y += y_change
time.tick(FPS)
pygame.quit()
quit()
答案 0 :(得分:1)
您可以使用collision response
rect2 = pygame.draw.rect(screen, green, [lead_x,lead_y, 50, 50])
rect1 = pygame.draw.rect(screen, blue, [lead_x1,lead_y1, 50 ,50])
if rect2.colliderect(rect1):
print("BOOM!")
如果你想要坐标:
print(rect2.left,rect2.right,rect2.top,rect2.bottom)
您可以使用here中的这些属性:
的x,y 顶部,左侧,底部,右侧 topleft,bottomleft,topright,bottomright midtop,midleft,midbottom,midright center,centerx,centery 大小,宽度,高度 W,H