如何通过按钮消除Kivy弹出窗口? [英] How to dismiss the Kivy pop-up via a Button?

问题描述

我用Kivy创建了一个弹出窗口，其中包含2个按钮.用户可以通过在弹出区域之外按下弹出窗口(auto_dismiss = True)或单击否"来关闭该弹出窗口. 选择是"按钮，将退出整个应用程序.

I have a pop-up created with Kivy, which contains 2 buttons. User can dismiss the pop-up by pressing outside of the pop-up area (auto_dismiss = True), or by clicking the "No" button. Selecting the "Yes" button, will exit the whole application.

请参阅相关代码:

class ExitApp(App):

def exit_confirmation(self):

    # popup can only have one Widget.  This can be fixed by adding a BoxLayout

    self.box_popup = BoxLayout(orientation = 'horizontal')

    self.box_popup.add_widget(Label(text = "Really exit?"))

    self.box_popup.add_widget(Button(
        text = "Yes",
        on_press = ExitApp.exit,
        size_hint = (0.215, 0.075)))

    self.box_popup.add_widget(Button(
        text = "No",
        on_press = self.popup_exit.dismiss,
        size_hint=(0.215, 0.075)))

    self.popup_exit = Popup(title = "Exit",
        content = self.box_popup,
        size_hint = (0.4, 0.4),
        auto_dismiss = True)

    self.popup_exit.open()

def exit(self):

    App.get_running_app().stop()

现在问题出在按下否"按钮.当按下该键时，代码退出，并显示以下错误:

The problem now lays with pressing the "No" button. When that is pressed, the code exits with this error:

 on_press = self.popup_exit.dismiss,

AttributeError:按钮"对象没有属性"popup_exit"

AttributeError: 'Button' object has no attribute 'popup_exit'

有什么主意，我如何才能尽可能轻松地解决此问题?

Any idea how I can fix this as easily as possible?

推荐答案

您可以通过惰性函数来解决此问题

You can solve this issue by a lazy function

on_press = lambda *args: self.popup_exit.dismiss()

这样，仅当按下按钮并且 popup_exit 已经就位时，查找才会仅进行.

This way, the lookup will occur only when the button is pressed and popup_exit is already in place...

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