Use loop to create x number of tkinter OptionMenu

Question

I am attempting to create an arbitrary number of optionmenus, but have trouble when trying to pull the StringVar() selected by each optionmenu.

Goal: Create an arbitrary number of optionmenus with consecutive names (up to that arbitrary number) and consecutive variable names keeping track of the current optiomenu value

For example if an optionmenu is as follows:

import tkinter as tk
from tkinter import *

root = tk.Tk()
root.geometry()

dropdown1 = StringVar()

Dropdownoptions = [
    "option1",
    "option2",
    "option3"
]

dropdownfirst = tk.OptionMenu(root, dropdown1, *Dropdownoptions)
dropdownfirst.grid(column=0, row=0)

root.mainloop()

If using a dictionary I do not know how to pull the values out of each Optionmenu. When looking at other questions about the use of dictionaries to create variables, most answers boil down to "Learn How to Use Dictionaries" instead of answering the questions.

There was a very similar problem posted in Tkinter Create OptionMenus With Loop but sadly it is not applicable in my case.

New code with grid and non-working button:

import tkinter as tk


def erase_option():
    for (name, var) in options.items():
        print(var.get())
        # print(options['optionmenu4'])
        # This just places label over dropdown, doesnt successfully take place for removal
        labelforemoval = tk.Label(text=" ")
        labelforemoval.grid(column=0, row=4)
        labelforemoval.grid_forget()


root = tk.Tk()

Dropdownoptions = [
    "option1",
    "option2",
    "option3"
]

maxval = 10

options = {}
for om, x in zip(range(maxval), range(maxval)):
    name = f"optionmenu{om}"
    var = tk.StringVar()
    options[name] = var
    name = tk.OptionMenu(root, var, *Dropdownoptions)
    name.grid(column=0, row=x)

button = tk.Button(root, text="Erase 5th option", command=erase_option)
button.grid(column=0, row=maxval)

root.mainloop()

score 2 · Accepted Answer · answered Dec 21 '22 at 23:38

Give each optionmenu its own StringVar instance. Save those instances in a list or dictionary. To get the values, iterate over the list.

The following code creates a dictionary named options. It creates a series of variables and optionmenus in a loop, adding each variable to this options dictionary. The function print_options iterates over the list printing out the key and value for each option.

import tkinter as tk

def print_options():
    for (name, var) in options.items():
        print(f"{name}: {var.get()}")

root = tk.Tk()

Dropdownoptions = [
    "option1",
    "option2",
    "option3"
]
options = {}
for om in range(10):
    name = f"Option {om}"
    var = tk.StringVar(value="")
    options[name] = var
    om = tk.OptionMenu(root, var, *Dropdownoptions)
    om.pack()

button = tk.Button(root, text="Print Options", command=print_options)
button.pack(side="bottom")

root.mainloop()

Edited question to include updated code. How would I remove a specific optionmenu from the grid? tk Doesn't support overwriting with other widgets and instead layers them. Thanks for the incredible help over the past few weeks — Luke-McDevitt, Dec 22 '22 at 18:37
@Luke-McDevitt: you would do it using the same technique. Save the option menus in a list or dictionary. — Bryan Oakley, Dec 22 '22 at 19:04

Use loop to create x number of tkinter OptionMenu

1 Answers1