kivy访问孩子id

Question

我想访问孩子的 id 来决定是否删除该小部件。我有以下代码：

主要.py

#!/usr/bin/kivy
# -*- coding: utf-8 -*-

from kivy.app import App
from kivy.uix.boxlayout import BoxLayout


class Terminator(BoxLayout):
    def DelButton(self):
        print("Deleting...")

        for child in self.children:
            print(child)
            print(child.text)

            if not child.id == 'deleto':
                print(child.id)
                #self.remove_widget(child)
            else:
                print('No delete')


class TestApp(App):
    def build(self):
        pass


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

测试.kv

#:kivy 1.9.0

<Terminator>:
    id: masta
    orientation: 'vertical'

    Button:
        id: deleto
        text: "Delete"
        on_release: masta.DelButton()

    Button
    Button

Terminator

但是，当使用：打印 id 时print(child.id)，它总是返回：None。即使print(child.text)正确返回Delete或 .

问题

• 为什么child.id不返回deleto，而是相反None？
• 如何检查是否删除带有“删除”的按钮，而是删除所有其他按钮？

Answer 1

正如您可以在文档中阅读的那样:

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在小部件树中，通常需要访问/引用其他小部件。Kv 语言提供了一种使用 id\xe2\x80\x99s 来执行此操作的方法。将它们视为只能在 Kv\n 语言中使用的类级别变量。

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这里描述了从 Python 代码访问 ids。工作示例：

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from kivy.app import App\nfrom kivy.uix.boxlayout import BoxLayout\nfrom kivy.lang import Builder\nfrom kivy.properties import ObjectProperty\nfrom kivy.uix.button import Button\n\nBuilder.load_string("""\n<Terminator>:\n    id: masta\n    orientation: \'vertical\'\n\n    MyButton:\n        id: deleto\n\n        button_id: deleto\n\n        text: "Delete"\n        on_release: masta.DelButton()\n\n    MyButton\n    MyButton\n""")\n\nclass MyButton(Button):\n    button_id = ObjectProperty(None)\n\nclass Terminator(BoxLayout):\n    def DelButton(self):\n        for child in self.children:\n            print(child.button_id)       \n\nclass TestApp(App):\n    def build(self):\n        return Terminator()    \n\nif __name__ == \'__main__\':\n    TestApp().run()\n

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要跳过删除带有“删除”标签的按钮，您可以检查其text property. Hovewer deleting from inside the loop will lead to the bugs, since some of the children will get skiped after the list you\'re iterating on will get altered:

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class Terminator(BoxLayout):\n    def DelButton(self):\n        for child in self.children:            \n            self.remove_widget(child) # this will leave one child\n

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您必须创建要删除的子项列表：

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class Terminator(BoxLayout):\n    def DelButton(self):\n        for child in [child for child in self.children]:            \n            self.remove_widget(child) # this will delete all children\n

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在你的情况下：

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class Terminator(BoxLayout):\n    def DelButton(self):\n        for child in [child for child in self.children if child.text != "Delete"]:            \n            self.remove_widget(child)\n

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