帶有可執行示例的多個復選框選擇中的簡單問題。是否有自動識別多個選中復選框的方法？

是否有一種自動方法來識別多個選中的復選框？我不想手動創建多個選擇組合。我想得到：

• 1/3 復選框：如果我單獨選擇第一個或第三個框，我會在文本框中列印“ok”
• 2/3 復選框：如果我同時選擇第一個和第三個，我會在文本框中得到列印“ok”。但是只列印了一個“ok”，而不是兩個 ok 為“ok ok”（所以兩個復選框對應一個 ok）
• 我不想在文本框中列印第二個復選框，因為我故意在上面寫了一個錯誤。

問題是這個功能。可能 AND（或 OR）不是我需要的，我猜，我不知道：

def aaa():
    if Checkbutton1.get() and Checkbutton2.get() and Checkbutton3.get():
        textbox.insert("end", "Ok")

#I also tried to write: if Button1_func() and Button2_func() and Button3_func():

我想解決方案是為每種型別的多選案例創建其他條件，但我認為有一種更自動和更好的方法，無需創建帶條件的選擇案例，因為我將不得不添加幾十個其他復選框，情況會變得復雜的。

重要提示：請不要對 True 和 False 進行諷刺，因為 5 3 = 8 只是一個簡單的示例，以了解我必須應用其他不同代碼的復選框的邏輯。保持原樣回傳 True 和 False，而不更改函式。

任何人都可以通過向我展示正確的代碼來幫助我嗎？謝謝

import sqlite3
from tkinter import *
from tkinter import ttk
import tkinter as tk
import tkinter.messagebox
from tkinter import messagebox

root = tk.Tk()
root.geometry("600x600")
root.configure(bg='white')

Checkbutton1 = IntVar()
Checkbutton2 = IntVar()
Checkbutton3 = IntVar()

#CHECKBOX'S FUNCTIONS
def Button1_func():
    if 5   3 == 8:
        return True
    else:
        return False

def Button2_func():
    if 5   3 == 7:
        return True
    else:
        return False

def Button3_func():
    if 5   3 == 8:
        return True
    else:
        return False

#CHECKBOX
Button1 = Checkbutton(root, text = "Checkbox 1", variable = Checkbutton1, onvalue = 1, offvalue = 0, height = 1,
                      bg="white", foreground='black', activebackground="white", command=Button1_func)
Button1.place(x=10, y=36)

Button2 = Checkbutton(root, text = "Checkbox 2", variable = Checkbutton2, onvalue = 1, offvalue = 0, height = 1,
                      bg="white", foreground='black', activebackground="white", command=Button2_func)
Button2.place(x=10, y=66)

Button3 = Checkbutton(root, text = "Checkbox 3", variable = Checkbutton3, onvalue = 1, offvalue = 0, height = 1,
                      bg="white", foreground='black', activebackground="white", command=Button3_func)
Button3.place(x=10, y=146)

#BUTTON FUNCTION
def aaa():
    if Checkbutton1.get() and Checkbutton2.get() and Checkbutton3.get():
        textbox.insert("end", "Ok")

    #I also tried to write: if Button1_func () and Button2_func () and Button3_func ():

#TEXTOBOX
textbox = tk.Text(root, width=33, height=10, font=('helvetic', 12))
textbox.place(x=30, y=220)

#BUTTON
button = tk.Button(root, text="Ok", command= lambda: [aaa()])
button.pack()


root.mainloop()

更新在代碼中還有這個我保存和加載復選框：

chk_lst = []
chk_lst.extend([Checkbutton1,Checkbutton2, Checkbutton3])

uj5u.com熱心網友回復：

您可以使用集合來存盤基于這些檢查按鈕狀態的函式參考。然后 insideaaa()執行集合中的所有函式，并確定是否將 OK 放入文本框中：

def clicked(flag, func):
    if flag:
        funclist.add(func)
    else:
        funclist.remove(func)

funclist = set()

#CHECKBOX
Button1 = tk.Checkbutton(root, text = "Checkbox 1", variable = Checkbutton1, onvalue = 1, offvalue = 0, height = 1,
                         bg="white", foreground='black', activebackground="white",
                         command=lambda: clicked(Checkbutton1.get(), Button1_func))
Button1.place(x=10, y=36)

Button2 = tk.Checkbutton(root, text = "Checkbox 2", variable = Checkbutton2, onvalue = 1, offvalue = 0, height = 1,
                         bg="white", foreground='black', activebackground="white",
                         command=lambda: clicked(Checkbutton2.get(), Button2_func))
Button2.place(x=10, y=66)

Button3 = tk.Checkbutton(root, text = "Checkbox 3", variable = Checkbutton3, onvalue = 1, offvalue = 0, height = 1,
                         bg="white", foreground='black', activebackground="white",
                         command=lambda: clicked(Checkbutton3.get(), Button3_func))
Button3.place(x=10, y=146)

#BUTTON FUNCTION
def aaa():
    # if need to clear the text box, uncomment below line
    #textbox.delete("1.0", "end")
    if funclist and all(func() for func in funclist):
        textbox.insert("end", "Ok")

uj5u.com熱心網友回復：

為澄清問題而編輯

因為每個Checkbutton都有一個與之關聯的獨特功能，所以您需要手動定義這些功能。現在，我已經替換了函式來代替。

您實際上不需要將函式傳遞給您的復選按鈕，因為它們僅在您單擊Button. 相反，您可以進行更改aaa，以便它為勾選的函式創建一個函式結果串列Checkbuttons。如果其中任何一個回傳False，請不要列印“OK”。但如果他們都回傳True，列印“OK”

import tkinter as tk

root = tk.Tk()
root.geometry("600x600")

# How many checkboxes you want
buttons = 5
button_eval = [0 for i in range(buttons)]

# If your functions are unique for each checkbox, you will need to define them here
def foo():
    return True
def bar():
    return False
my_functions = [foo, bar, foo, bar, foo] # You need to have 5 functions in this case for this to work

# Creating variables, buttons etc
x = [10, 10, 10, 10, 10]
y = [36, 66, 96, 126, 156]
checkbuttons_vars = [tk.IntVar() for _ in range(buttons)]
checkbuttons = [tk.Checkbutton(root, text=f"Checkbox {i 1}", variable=checkbuttons_vars[i], onvalue=1, offvalue=0) for i in range(buttons)]
for i, box in enumerate(checkbuttons):
    box.place(x=x[i],y=y[i])

#BUTTON FUNCTION
def aaa():
    function_results = [function() for i, function in enumerate(my_functions) if checkbuttons_vars[i].get()]
    textbox.delete(1.0, "end")
    if False not in function_results:
        textbox.insert(1.0, "Ok")

#TEXTOBOX
textbox = tk.Text(root, width=33, height=10, font=('helvetic', 12))
textbox.pack()

#BUTTON
button = tk.Button(root, text="Ok", command=aaa)
button.pack()

root.mainloop()

標籤：Python python-3.x tkinter 复选框

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