如何与 Kivy 中的标签发生碰撞

Question

我有这段代码，但我不知道如何碰撞 Label: Tag 和 Buttons。这是代码：

 def _on_keyboard_down(self, keyboard, keycode, text, modifiers):
        if keycode[1] == 'z' or keycode[1] == 'up':
            self.pY += .1
            for wid in self.walk():
                if isinstance(wid, Button):
                    if Tag.collide_widget(wid):

所以，我 walk() 主要 class，如果它是一个按钮，我使用 collide_widget。我想无论我尝试将什么作为函数的参数，似乎都不起作用。

这里是完整的代码：**主文件**

from kivy.app import App
from kivy.config import Config
from kivy.uix.floatlayout import FloatLayout
from kivy.properties import NumericProperty
from kivy.core.window import Window
from kivy.uix.label import Label
from kivy.uix.button import Button

Window.size = (900, 600)
Config.set('graphics', 'resizable', True)


class Tag(Label):
    pass


class MainWindow(FloatLayout):
    pY = NumericProperty(.6)
    pX = NumericProperty(.6)

    def __init__(self, **kwargs):

        super(MainWindow, self).__init__(**kwargs)
        self._keyboard = Window.request_keyboard(self._keyboard_closed, self)
        self._keyboard.bind(on_key_down=self._on_keyboard_down)

    def _keyboard_closed(self):
        self._keyboard.unbind(on_key_down=self._on_keyboard_down)
        self._keyboard = None

    def _on_keyboard_down(self, keyboard, keycode, text, modifiers):
        if keycode[1] == 'z' or keycode[1] == 'up':
            self.pY += .1
            for wid in self.walk():
                if isinstance(wid, Button):
                    if Tag.collide_widget(Tag, wid):
                        self.pY -= .2
        elif keycode[1] == 's' or keycode[1] == 'down':
            self.pY -= .1
        elif keycode[1] == 'q' or keycode[1] == 'left':
            self.pX -= .1
        elif keycode[1] == 'd' or keycode[1] == 'right':
            self.pX += .1
        elif keycode[1] == 'o':
            return 0
        return True


class FenetreApp(App):
    def build(self):
        return MainWindow()


FenetreApp().run()

Fenetre.kv 文件:

<Button>:
     size_hint: 0.1, 0.1
     background_color: 0.1, 0.5, 0.6, 1


<Tag@Label>:
     size_hint: 0.1, 0.1
     background_color: 1, 0, 0, 1
     canvas.before:
          Color:
               rgb: 0.1, 0.6, 0
          Rectangle:
               pos: self.pos
               size: self.size

<MainWindow@FloatLayout>:
     Button:
          text: "Up"
          pos_hint: {"x":0.8, "top":1}
          on_press: self.parent.pY= self.parent.pY +0.1
     Button:
          text: "Down"
          pos_hint: {"x":0.8, "top":0.8}
          on_press: self.parent.pY= self.parent.pY -0.1
     Button:
          text: "Left"
          pos_hint: {"x":0.7, "top":0.9}
          on_press: self.parent.pX = self.parent.pX -0.1
     Button:
          text: "Right"
          pos_hint: {"x":0.9, "top":0.9}
          on_press: self.parent.pX = self.parent.pX +0.1
     Tag:
          name: "L1"
          text: "move"
          pos_hint: {"x":self.parent.pX, "top":self.parent.pY}

Answer 1

if Tag.collide_widget(Tag, wid):

这没有意义，Tag 是 class 定义本身，是 Python 代码中的抽象，与您应用中的任何特定元素无关。您需要在您的应用中调用 collide_widget 以获取此 class 的特定实例。

Answer 2

所以，我找到了解决问题的方法：向小部件添加一个 id：

Tag:
          id: Tag_Layout
          name: "L1"
          text: "move"
          pos_hint: {"x":self.parent.pX, "top":self.parent.pY}

然后通过id访问对象：

for wid in self.walk():
            if isinstance(wid, Button):
                if self.ids.Tag_Layout.collide_widget(wid):
                    self.pY -= .2

如何与 Kivy 中的标签发生碰撞

How to do collision with a Label in Kivy

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collision-detection

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kivy-language