Question

逻辑：

我按下窗口上的按钮，以打开第一个弹出窗口；

我在第一个弹出窗口上按下一个按钮，以打开第二个弹出窗口；

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

这是示例代码：

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“第一个弹出窗口”上的“标签”文本未更改。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”

我无法使其工作。

Answer 1

在您的open_first_popup()方法中，使用self.the_first_popup = FirstPopup()。这样会将the_first_popup保存为实例变量，您以后可以在self.the_first_popup类中以FirstScreen的形式访问。

因此，这是您代码的修改版本：

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()

我已删除了access_first_screen，因为这只是创建了一个FirstScreen的新实例，该实例与您的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（GUI中的实例确实有这样的属性）。我还消除了access_first_popup。

我也在上面包括了我的建议。

Kivy-在子弹出窗口上按一个按钮可以更改父弹出窗口上的标签文本

1 个答案: