Kivy-按下子弹出窗口上的按钮以更改父弹出窗口上的标签文本

Question

逻辑：

我按下窗口上的一个按钮，打开第一个弹出窗口；

我在第一个 Popup 上按下一个按钮，打开第二个 Popup；

我按下第二个 Popup 上的按钮，调用窗口类中的函数，更改第一个 Popup 上的标签文本。

这里是示例代码：

from kivy.app import App
from kivy.lang import Builder
from kivy.uix.screenmanager import ScreenManager, Screen
from kivy.uix.popup import Popup
from kivy.uix.label import Label
from kivy.uix.button import Button
Builder.load_string("""
<FirstScreen>:
    BoxLayout:
        Button:
            text: 'open the first Popup'
            on_release: root.open_first_popup()
        Button:
            text: 'Quit'

<FirstPopup>:
    first_label: first_label
    BoxLayout:
        orientation: 'vertical'
        Label:
            id: first_label
            text: 'text on first Popup'
        Button:
            text: 'open second Popup'
            on_release: app.access_first_screen.open_second_popup()

<SecondPopup>:
    BoxLayout:
        orientation: 'vertical'
        Label:
            id: second_label
            text: 'text on second Popup'
        Button:
            text: 'change label text on first Popup'
            on_release: app.access_first_screen.change_label_text_on_first_popup()
        Button:
            text: 'close me to check if label text on first Popup has changed'
            on_release: root.close_me()

""")

class FirstPopup(Popup):
    pass

class SecondPopup(Popup):
    def __init__(self, **kwargs):
        super(SecondPopup, self).__init__(**kwargs)

    def close_me(self):
        self.dismiss()

class FirstScreen(Screen):
    def open_first_popup(self):
        the_first_popup = FirstPopup()
        the_first_popup.open()

    def open_second_popup(self):
        the_second_popup = SecondPopup()
        the_second_popup.open()

    def change_label_text_on_first_popup(self):
        app = App.get_running_app()
        app.access_first_screen.the_first_popup.first_label.text = 'change to me'

class Root(ScreenManager):
    def __init__(self, **kwargs):
        super(Root, self).__init__(**kwargs)
        xx = FirstScreen(name='first screen')
        self.add_widget(xx)

class TestApp(App):
    access_first_screen = FirstScreen()
    access_first_popup = FirstPopup()
    def build(self):
        return Root()

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

现在的错误是：

AttributeError: 'FirstScreen' 对象没有属性 'the_first_popup'

然后我尝试了：

def change_label_text_on_first_popup(self):
    app = App.get_running_app()
    app.access_first_popup.first_label.text = 'change to me'

没有错误，但第一个 Popup 上的 Label 文本没有改变。 然后我尝试了：

def change_label_text_on_first_popup(self):
    app = App.get_running_app()
    app.ids.first_label.text = 'change to me'

错误是： AttributeError: 'TestApp' 对象没有属性 'ids'

然后我尝试了：

def change_label_text_on_first_popup(self):
    app = App.get_running_app()
    self.the_first_popup.first_label.text = 'change to me'

错误是： AttributeError：“FirstScreen”对象没有属性“the_first_popup”

我无法让它工作。

Answer 1

在您的open_first_popup() 方法中使用self.the_first_popup = FirstPopup()。 这会将the_first_popup 保存为一个实例变量，您以后可以在FirstScreen 类中以self.the_first_popup 的形式访问它。

所以，这是您的代码的修改版本：

from kivy.app import App
from kivy.lang import Builder
from kivy.uix.screenmanager import ScreenManager, Screen
from kivy.uix.popup import Popup

Builder.load_string("""
#:import Factory kivy.factory.Factory

<FirstScreen>:
    BoxLayout:
        Button:
            text: 'open the first Popup'
            on_release: root.open_first_popup()
        Button:
            text: 'Quit'

<FirstPopup>:
    first_label: first_label
    BoxLayout:
        orientation: 'vertical'
        Label:
            id: first_label
            text: 'text on first Popup'
        Button:
            text: 'open second Popup'
            on_release: Factory.SecondPopup().open()

<SecondPopup>:
    BoxLayout:
        orientation: 'vertical'
        Label:
            id: second_label
            text: 'text on second Popup'
        Button:
            text: 'change label text on first Popup'
            on_release: app.root.get_screen("first screen").change_label_text_on_first_popup()
        Button:
            text: 'close me to check if label text on first Popup has changed'
            on_release: root.close_me()
""")

class FirstPopup(Popup):
    pass

class SecondPopup(Popup):
    def __init__(self, **kwargs):
        super(SecondPopup, self).__init__(**kwargs)

    def close_me(self):
        self.dismiss()

class FirstScreen(Screen):
    def open_first_popup(self):
        self.the_first_popup = FirstPopup()
        self.the_first_popup.open()

    def open_second_popup(self):
        the_second_popup = SecondPopup()
        the_second_popup.open()

    def change_label_text_on_first_popup(self):
        self.the_first_popup.first_label.text = 'change to me'

class Root(ScreenManager):
    def __init__(self, **kwargs):
        super(Root, self).__init__(**kwargs)
        xx = FirstScreen(name='first screen')
        self.add_widget(xx)

class TestApp(App):
    def build(self):
        return Root()

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

我已经删除了access_first_screen，因为这只会创建一个新的FirstScreen 实例，它与您的GUI 中出现的实例无关。这就是为什么它总是说：

'FirstScreen' object has no attribute 'the_first_popup'

因为FirstScreen 的新实例没有这样的属性（您的GUI 中的实例确实有这样的属性）。我还删除了access_first_popup。

我在上面也包含了我的建议。