我想制作一个Python游戏,其中红龟追逐蓝龟。当红龟抓到蓝龟时,我想让它在屏幕上说'COLLISION',但它不起作用。当它碰撞时,没有任何反应,它给了我一个错误'乌龟'对象不可调用'。在Python龟游戏中检测碰撞
from turtle import Turtle, Screen
playGround = Screen()
playGround.screensize(250, 250)
playGround.title("Turtle Keys")
run = Turtle("turtle")
run.speed("fastest")
run.color("blue")
run.penup()
run.setposition(250, 250)
follow = Turtle("turtle")
follow.speed("fastest")
follow.color("red")
follow.penup()
follow.setposition(-250, -250)
def k1():
run.forward(45)
def k2():
run.left(45)
def k3():
run.right(45)
def k4():
run.backward(45)
def quitThis():
playGround.bye()
def follow_runner():
follow.setheading(follow.towards(run))
follow.forward(8)
playGround.ontimer(follow_runner, 10)
playGround.onkey(k1, "Up") # the up arrow key
playGround.onkey(k2, "Left") # the left arrow key
playGround.onkey(k3, "Right") # you get it!
playGround.onkey(k4, "Down")
playGround.listen()
follow_runner()
def is_collided_with(self, run):
return self.rect.colliderect(run.rect)
runner = run(10, 10, 'my_run')
follower = follow(20, 10)
if follow.is_collided_with(run):
print 'collision!'
playGround.mainloop()
也许你可以尝试检查'run'的位置是否与'follow'匹配? – aug
我在关于pygame的代码中看不到任何东西。如果你正在使用pygame,为什么不使用它的精灵功能,让pygame检测碰撞? –