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Question

我正在编写一个程序，它需要更改标签和将文本输入文本框的功能，但我不知道如何使用我正在使用的新编程方式（面向对象）。我之前完成了程序，但是使用此代码生成了框架：

f = [Frame(root) for i in range(0,5)]
for i in f:
    i.place(relx=0,rely=0,relwidth=1,relheight=1)

然后我把它全部放在我跑的一个班级中，但那是一个糟糕的形式，所以我重做它。到目前为止，我的代码如下：

import tkinter as tk
import datetime,time,os,sys
import sqlite3 as lite
from datetime import date

Title_Font= ("Times", 18, "underline italic")

unix = time.time()
time_curr = str(datetime.datetime.fromtimestamp(unix).strftime('%H:%M'))
date1 = str(datetime.datetime.fromtimestamp(unix).strftime('%d-%m-%Y'))


class Creating_Stuff(tk.Tk):

    def __init__(self, *args, **kwargs):

        tk.Tk.__init__(self, *args, **kwargs)
        container = tk.Frame(self)


        container.pack(side="top", fill="both", expand = True)

        container.grid_rowconfigure(0, weight=1)
        container.grid_columnconfigure(0, weight=1)

        self.frames = {}

        for F in (StartPage, PageOne, PageTwo):

            frame = F(container, self)

            self.frames[F] = frame

            frame.grid(row=0, column=0, sticky="nsew")

        self.show_frame(StartPage)

    def show_frame(self, cont):

        frame = self.frames[cont]
        frame.tkraise()

class StartPage(tk.Frame):

    def __init__(self, parent, controller):
        tk.Frame.__init__(self,parent)
        self.grid_columnconfigure(0, weight=1)
        self.grid_columnconfigure(3, weight=1)       

        self.option_add( "*font", "Times 12" )
        self.tk_setPalette(background='#bbfff0', foreground='black',
               activeBackground='#d9d9d9', activeForeground='#ff9933')


        label = tk.Label(self, text="Laptop Booking System", font=Title_Font)
        label.grid(row=0, column=1, pady = 30)

        time = tk.Label(self, text="Time: " + time_curr + "\nDate: " + date1, font="Times 10")
        time.grid(row=0, column=2,columnspan=2)

        Booking_1 = tk.Button(self, text="Booking",
                            command=lambda: controller.show_frame(PageOne),bg='#f2f2f2',width=20)
        Booking_1.grid(row=2, column=1, pady=10)

        Tbl_q1 = tk.Button(self, text="Table Querying",
                            command=lambda: controller.show_frame(PageTwo),bg='#f2f2f2',width=20)
        Tbl_q1.grid(row=3, column=1, pady=10)

        Exit = tk.Button(self, text ="Exit",command=lambda:destroy(),bg='#f2f2f2')
        Exit.grid(row=5, column=1)


class PageOne(tk.Frame):

    def __init__(self, parent, controller):
        tk.Frame.__init__(self, parent)
        self.grid_columnconfigure(0, weight=1)
        self.grid_columnconfigure(3, weight=1)
        label = tk.Label(self, text="Page One!!!", font=Title_Font)
        label.grid(row=1, column=1)


        bk2_menu = tk.Button(self, text="Back to Home",
                            command=lambda: controller.show_frame(StartPage))
        bk2_menu.grid(row=3, column=1)



class PageTwo(tk.Frame):

    def __init__(self, parent, controller):
        tk.Frame.__init__(self, parent)
        self.grid_columnconfigure(0, weight=1)
        self.grid_columnconfigure(3, weight=1)
        label = tk.Label(self, text="Page Two!!!", font=Title_Font)
        label.grid(row=1, column=1)


        bk2_Menu2 = tk.Button(self, text="Back to menu",
                            command=lambda: controller.show_frame(StartPage))
        bk2_Menu2.grid(row=3, column=1)



app = Creating_Stuff()
def destroy():
    app.destroy()

app.title("Laptop Booking System")
app.geometry("700x400")
app.mainloop()

如果你尝试一下它可以工作，但它只有3个不同的帧。如何在一个框架中获得一个按钮，使其在按下后在框架中显示“Hello”？

Answer 1

我修改了您的一个网页类，以说明如何完成。它包括添加一个Label来保存邮件，一个Button来控制它，一个函数，简称handler()，在按下后者时调用。它通过使小部件包含Frame子类实例self的属性来保存小部件，因此可以在handler()函数中轻松引用它们（无需借助全局变量）。

class PageOne(tk.Frame):
    def __init__(self, parent, controller):
        tk.Frame.__init__(self, parent)
        self.grid_columnconfigure(0, weight=1)
        self.grid_columnconfigure(3, weight=1)
        label = tk.Label(self, text="Page One!!!", font=Title_Font)
        label.grid(row=1, column=1)

        self.msg_label = tk.Label(self, text="")
        self.msg_label.grid(row=2, column=1)
        self.msg_button = tk.Button(self, text='Show Message',
                                    command=self.handler)
        self.msg_button.grid(row=3, column=1)
        self.msg_toggle = False

        bk2_menu = tk.Button(self, text="Back to Home", 
                             command=lambda: controller.show_frame(StartPage))
        bk2_menu.grid(row=5, column=1)

    def handler(self):
        self.msg_toggle = not self.msg_toggle
        if self.msg_toggle:
            self.msg_label.config(text='Hello')
            self.msg_button.config(text='Clear Message')
        else:
            self.msg_label.config(text='')
            self.msg_button.config(text='Show Message')

截图

按下按钮前：

按下按钮后：

面向对象的Tkinter函数 - 我如何将它放入？

1 个答案:

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