我正在尝试让我的绿色方 block 检测与蓝色的碰撞 square,但是,我不知道如何构造 if 语句,使其在接触时立即发生碰撞。
如果我使 rect1 的位置 >= 200,100
如果越过蓝色方 block 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()
最佳答案
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)
这些是您可以使用的属性,取自 here :
x,y top, left, bottom, right topleft, bottomleft, topright, bottomright midtop, midleft, midbottom, midright center, centerx, centery size, width, height w,h
关于python - pygame,如何获取图形的坐标?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31211582/