我想製作一個Python遊戲,其中紅龜追逐藍龜。當紅龜抓到藍龜時,我想讓它在屏幕上說'COLLISION',但它不起作用。當它碰撞時,沒有任何反應,它給了我一個錯誤'烏龜'對象不可調用'。在Python龜遊戲中檢測碰撞
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()
也許你可以嘗試檢查'run'的位置是否與'follow'匹配? – aug
我在關於pygame的代碼中看不到任何東西。如果你正在使用pygame,爲什麼不使用它的精靈功能,讓pygame檢測碰撞? –