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python - 使用 python/kivy 运行应用程序时仅显示“else 语句”

问题描述

我一直在使用 Kivy 开发一个应用程序，该应用程序接收 0、1 或 2 作为用户输入的文本，最终屏幕上的标签显示的文本应该根据接收到的输入而有所不同。它不再出现任何错误，但即使满足“if 语句”的要求，最终屏幕也只会显示“else”文本（“Bad Luck”）。

通过搜索类似的问题，我尝试使用整数的“input_filter”来解决问题，但它没有。

我是一个没有编程背景的初学者，所以请随意在你的答案中尽可能详细。提前感谢您的时间和您可能分享的任何想法/建议。以下是 .py 和 .kv 文件的代码：

.py 文件：

import kivy
from kivy.app import App
from kivy.lang import Builder
from kivy.uix.label import Label
from kivy.uix.gridlayout import GridLayout
from kivy.uix.textinput import TextInput
from kivy.uix.button import Button
from kivy.uix.widget import Widget
from kivy.uix.screenmanager import ScreenManager, Screen
from kivy.properties import NumericProperty, StringProperty, NumericProperty


class MainWindow(Screen):
    pass

class SecondWindow(Screen):
    pass

class ThirdWindow(Screen):
    pass

class FourthWindow(Screen):
    pass
    
    
class WindowManager(ScreenManager):
    pass


class MyApp(App):
    def build(self):
       
        kv = Builder.load_file("my.kv")
        return kv
        
     
    
if __name__=="__main__":
    MyApp().run()

.kv 文件：

WindowManager:
    MainWindow:
    SecondWindow:
    ThirdWindow:
    FourthWindow:
    

<MainWindow>:
    name: "main"
    GridLayout:
        cols:1
        rows:2
        Label:
            text: "stuff"
        Button:
            text: "stuff"
            on_release:
                app.root.current = "second"
                root.manager.transition.direction = "left"

<SecondWindow>:
    name: "second"
    GridLayout:
        cols:1
        rows:2

        GridLayout:
            cols:2
            Label:
                text: "stuff"
            TextInput:
                id: ti_a
                multiline:False
                
                
            Label:
                text: "stuff"
            TextInput:
                id: ti_b
                multiline:False
                

            Label:
                text: "stuff"
            TextInput:
                id: ti_c
                multiline:False
                

            Label:
                text: "stuff"
            TextInput:
                id: ti_d
                multiline:False
                

            Label:
                text: "stuff"
            TextInput:
                id: ti_e
                multiline:False
                

        GridLayout:
            cols:2 
        
            Button:
                text: "stuff"
                on_release:
                    app.root.current = "third"
                    root.manager.transition.direction = "left"
            Button:
                text: "Back"
                on_release:
                    app.root.current = "main"
                    root.manager.transition.direction = "right"

<ThirdWindow>:
    name: "third"
    GridLayout:
        cols:1
        rows:2

        GridLayout:
            cols:2
            Label:
                text: "stuff"
            TextInput:
                id: ti_f
                multiline:False
                

        GridLayout:
            cols:2

            Button:
                text: "stuff"
                on_release:
                    app.root.current = "fourth"
                    root.manager.transition.direction = "left"

            Button:
                text: "Back"
                on_release:
                    app.root.current = "second"
                    root.manager.transition.direction = "right"


<FourthWindow>:
    name: "fourth"
    Label:
        text: "Stuff" if root.manager.get_screen("second").ids.ti_a.text == "0" and root.manager.get_screen("second").ids.ti_b.text == "0" and root.manager.get_screen("second").ids.ti_c.text == "0" and root.manager.get_screen("second").ids.ti_d.text == "0" and root.manager.get_screen("second").ids.ti_e.text == "1" and root.manager.get_screen("third").ids.ti_f.text == "0" else "Bad Luck"

标签： pythonif-statementkivy

解决方案

问题是 kivy 无法识别表达式：

root.manager.get_screen("second").ids.ti_a.text

作为它可以绑定的属性，因此Label您的 text的值FourthWindow仅被评估一次（在FourthWindow创建时）。在您的应用程序中对另一个进行任何更改Labels都不会影响FourthWindow. 请参阅文档。

解决方法是编写您自己的代码，Label通过添加一个id来更新它Label：

<FourthWindow>:
    name: "fourth"
    Label:
        id: lab
        text: ""

并添加一个on_pre_enter()方法FourthWindow：

class FourthWindow(Screen):
    def on_pre_enter(self, *args):
        if self.manager.get_screen("second").ids.ti_a.text == "0" and \
                self.manager.get_screen("second").ids.ti_b.text == "0" and \
                self.manager.get_screen("second").ids.ti_c.text == "0" and \
                self.manager.get_screen("second").ids.ti_d.text == "0" and \
                self.manager.get_screen("second").ids.ti_e.text == "1" and \
                self.manager.get_screen("third").ids.ti_f.text == "0":
            self.ids.lab.text = 'Stuff'
        else:
            self.ids.lab.text = 'Bad Luck'

这将评估您的表达式并在显示text之前设置每次FourthWindow。

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