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Set a default value for a ttk Combobox

I'm using Python 3.2.1 in Arch Linux x86_64. This one is really driving me crazy: I just want to have a default, preselected value for a ttk.Combobox as soon as I grid it. This is my code:

from tkinter import Tk, StringVar, ttk

root = Tk()

def combo(parent):
    value = StringVar()
    box = ttk.Combobox(parent, textvariable=value, state='readonly')
    box['values'] = ('A', 'B', 'C')
    box.current(0)
    box.grid(column=0, row=0)

combo(root)

root.mainloop()

Which draws an empty Combobox. What's funny is that if I don't use a function it works perfectly:

from tkinter import Tk, StringVar, ttk

root = Tk()

value = StringVar()
box = ttk.Combobox(root, textvariable=value, state='readonly')
box['values'] = ('A', 'B', 'C')
box.current(0)
box.grid(column=0, row=0)

root.mainloop()

Of course, in the real program I have to use a function, so I need another solution.

like image 630
kynikos Avatar asked Jul 29 '11 17:07

kynikos


2 Answers

The problem is that the instance of StringVar is getting garbage-collected. This is because it's a local variable due to how you wrote your code.

One solution is to use a class so that your StringVar persists:

from tkinter import Tk, StringVar, ttk

class Application:

    def __init__(self, parent):
        self.parent = parent
        self.combo()

    def combo(self):
        self.box_value = StringVar()
        self.box = ttk.Combobox(self.parent, textvariable=self.box_value, 
                                state='readonly')
        self.box['values'] = ('A', 'B', 'C')
        self.box.current(0)
        self.box.grid(column=0, row=0)

if __name__ == '__main__':
    root = Tk()
    app = Application(root)
    root.mainloop()
like image 164
Eryk Sun Avatar answered Oct 03 '22 07:10

Eryk Sun


When your function 'combo' exits, the local variable 'value' is destroyed. You need a persistent variable, such as a global variable or a variable that is a property of a class so that the value isn't garbage-collected while the widget still exists.

like image 38
Bryan Oakley Avatar answered Oct 03 '22 06:10

Bryan Oakley