tkinter:lambda多个按钮到一个标签? [英] tkinter: lambda multiple Buttons to a Label?

问题描述

说明:我希望能够单击一个按钮，并将其值发送到标签小部件(lab_1)，我知道要做一个按钮，您必须使用command=lambda: lab_1.configure(text=0)，但是如何当我使用for循环在嵌入式列表中创建所有按钮时，就可以将值打印到标签上.截至目前，所有按钮的变量名为key_press.

Description: I would like to be able to click a button and have it send it's value to the label widget which is (lab_1), I understand to do one single button you have to use command=lambda: lab_1.configure(text=0) but how would you go about printing values to the label when I have used a for loop to create all of my buttons in an embedded list. as of right now all of the buttons have the variable named key_press.

简化问题:

• 如何将所有值写入我的标签(lab_1)?

如何使用for循环，因此每次单击该按钮时，它将在Label中显示其值?

How can you use a for loop so every time that button is clicked it will display its value to the Label?

import tkinter as tk

root = tk.Tk()
# Change the original tkinter icon
root.iconbitmap('C:\\Users\\bmxfi\Downloads\images.ico')
# Set the title of the window
root.title('Calculator')
# Change the geometry of the window
root.geometry('300x450+750+150')
# Setting Minimum and Maximum window width and height
root.minsize(width=300, height=450)
root.maxsize(width=300, height=450)
# Create Rows and Columns to display widgets
root.rowconfigure(0, weight=100)
root.rowconfigure(1, weight=100)
root.rowconfigure(2, weight=100)
# Columns start here:
root.columnconfigure(0, weight=100)
root.columnconfigure(1, weight=100)
root.columnconfigure(2, weight=100)
# Configuring the Main window named: root
root.configure(bg='#353535')
# Creating the main label that will display the calculated results
lab_1 = tk.Label(root, width=40)
lab_1.grid(row=0, column=1, columnspan=1, sticky='we')
lab_1.configure(bg='#545454', font=('Computer Modern', 25, 'bold'),
                fg='White', justify='right')
# Main Calculator Layout
# List Box for a calculator layout
box_1 = tk.Listbox(root)
box_1.grid(row=1, column=1, rowspan=2)
box_1.configure(bg='#353535', relief='flat')
# Button Layout
calculator_layout = [[(7, 1), (8, 1), (9, 1), ('(', 1), (')', 1)],
                     [(4, 1), (5, 1), (6, 1), ('×', 1), ('÷', 1)],
                     [(1, 1), (2, 1), (3, 1), ('+', 1), ('-', 1)],
                     [(0, 1), ('±', 1), ('.', 1), ('-', 1), ('=', 1)]]

# Iterating through my list to get a set of buttons
row = 0
for i in calculator_layout:
    col = 0
    for a in i:
        key_press = tk.Button(box_1, text=a[0])
        key_press.grid(row=row, column=col, ipadx=10, ipady=8, sticky='we')
        key_press.configure(background='#545454', fg='white',
                            font=('Computer Modern', 10),
                            activebackground='red', state='normal')
        col += a[1]
    row += 1

root.mainloop()

推荐答案

您可以通过为lambda每个Button提供一个具有默认值的参数来实现.

You can do it by giving the lambda each Button an argument with a default value. There's no need for any extra for loops doing it this way (it's done in the one that creates the Buttons).

如何执行该操作，下面带有# CHANGED注释的地方显示.每个lambda函数中定义的其余代码均通过调用其configure()方法并将其传递给text选项的当前内容，从而重新配置名为lab_1的Label的text选项.关联的Button的符号.

How to do that is shown below in the spot with the # CHANGED comment. The rest of the code in each lambda function defined each re-configures the text option of the Label named lab_1 by calling its configure() method and passing it the current contents of the text option with the symbol of the associated Button appended to it.

...

# Iterating through my list to get a set of buttons
row = 0
for i in calculator_layout:
    col = 0
    for a in i:
        key_press = tk.Button(box_1, text=a[0])
        key_press.grid(row=row, column=col, ipadx=10, ipady=8, sticky='we')
        key_press.configure(background='#545454', fg='white',
                            font=('Computer Modern', 10),
                            activebackground='red', state='normal',
                            # CHANGED.
                            command=lambda sym=str(a[0]):
                                lab_1.configure(text=lab_1.cget('text')+sym))
        col += a[1]
    row += 1

root.mainloop()

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