gpt4 book ai didi

python - 如何在kivy Pong球游戏中从另一个类调用一个类的函数

转载 作者:行者123 更新时间:2023-12-01 05:09:41 25 4
gpt4 key购买 nike

我正在使用教程中给出的 PongGame 代码练习 Kivy。我想知道如何从新创建的类 - PongSample 中调用 PongGame 类中的函数 -serve_ball2()。在下面的代码中,我创建了一个 PongSample 类,以便在第一个球与 Racket 碰撞时发给第二个球。

更新:我可以从 PongSample 调用serve_ball2(),但serve_ball2() 无法按预期运行,即它不发球。

我在下面分享了完整的代码。提前致谢

Pong.py:

from kivy.app import App
from kivy.uix.widget import Widget
from kivy.properties import NumericProperty, ReferenceListProperty,\
ObjectProperty
from kivy.vector import Vector
from kivy.clock import Clock, time
from threading import Thread

class PongPaddle(Widget):
score = NumericProperty(0)

def bounce_ball(self, ball):
if self.collide_widget(ball):
vx, vy = ball.velocity
offset = (ball.center_y - self.center_y) / (self.height / 2)
bounced = Vector(-1 * vx, vy)
vel = bounced * 1.1
ball.velocity = vel.x, vel.y + offset
PongSample().call_game()

class PongBall(Widget):
velocity_x = NumericProperty(0)
velocity_y = NumericProperty(0)
velocity = ReferenceListProperty(velocity_x, velocity_y)

def move(self):
self.pos = Vector(*self.velocity) + self.pos

class PongSample(Widget):
def call_game(self):
print 'PongSample'
ponggame=PongGame()
ponggame.serve_ball2()

class PongGame(Widget):
ball = ObjectProperty(None)
ball2 = ObjectProperty(None)
player1 = ObjectProperty(None)
player2 = ObjectProperty(None)

def serve_ball(self, vel=(4, 0)):
self.ball.center = self.center
self.ball.velocity = vel

def serve_ball2(self, vel=(3, 0)):
print 'Serve_ball2'
self.ball2.center = self.center
self.ball2.velocity = vel

def serve_down(self):
print 'Inside Serve Down'
self.ball.center = self.center
self.ball.velocity = Vector(4,0).rotate(-90)

def update(self, dt):
self.ball.move()
self.ball2.move()

#bounce of paddles
self.player1.bounce_ball(self.ball)
self.player2.bounce_ball(self.ball)

#bounce ball off bottom or top
if (self.ball.y < self.y) or (self.ball.top > self.top):
self.ball.velocity_y *= -1
if (self.ball2.y < self.y) or (self.ball2.top > self.top):
self.ball2.velocity_y *= -1

#went of to a side to score point?
if self.ball.x < self.x:
self.player2.score += 1
self.serve_ball(vel=(4, 0))
if self.ball.x > self.width:
self.player1.score += 1
self.serve_ball(vel=(-4, 0))

if self.ball2.x < self.x:
self.player2.score += 1
self.serve_ball2(vel=(3, 0))
if self.ball2.x > self.width:
self.player1.score += 1
self.serve_ball2(vel=(-3, 0))

def on_touch_move(self, touch):
if touch.x < self.width / 3:
self.player1.center_y = touch.y
if touch.x > self.width - self.width / 3:
self.player2.center_y = touch.y

class PongApp(App):
def build(self):
game = PongGame()
game.serve_ball()
Clock.schedule_interval(game.update, 1.0 / 60.0)
return game

if __name__ == '__main__':
PongApp().run()

pong.kv:

    #:kivy 1.8.0

<PongBall>:
size: 50, 50
canvas:
Ellipse:
pos: self.pos
size: self.size

<PongPaddle>:
size: 25, 200
canvas:
Rectangle:
pos:self.pos
size:self.size

<PongGame>:
ball: pong_ball
ball2: pong_ball2
player1: player_left
player2: player_right

canvas:
Rectangle:
pos: self.center_x-5, 0
size: 10, self.height

Label:
font_size: 70
center_x: root.width / 4
top: root.top - 50
text: str(root.player1.score)

Label:
font_size: 70
center_x: root.width * 3 / 4
top: root.top - 50
text: str(root.player2.score)

PongBall:
id: pong_ball
center: self.parent.center

PongBall:
id: pong_ball2
center: self.parent.center

PongPaddle:
id: player_left
x: root.x
center_y: root.center_y

PongPaddle:
id: player_right
x: root.width-self.width
center_y: root.center_y

最佳答案

要使用您的类,请将 game.serve_ball2() 添加到 PongApp

class PongApp(App):
def build(self):
game = PongGame()
game.serve_ball()
game.serve_ball2()
Clock.schedule_interval(game.update, 1.0 / 60.0)
return game

并添加self.ball2以从 Racket 上弹起:

#bounce of paddles
self.player1.bounce_ball(self.ball)
self.player2.bounce_ball(self.ball)
self.player1.bounce_ball(self.ball2)
self.player2.bounce_ball(self.ball2)

关于python - 如何在kivy Pong球游戏中从另一个类调用一个类的函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24424962/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com