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I'm learning to code a program with GUI, but I can't get my head around with the best principle.

At the moment I'm trying to have 5 entries and the text which would be written to entries would automatically update to labels. Here is my code so far:

import tkinter as tk


class MainApplication(tk.Frame):
    def __init__(self, parent, *args, **kwargs):
        tk.Frame.__init__(self, parent, *args, **kwargs)
        self.parent = parent
    
        # GUI starts here

        self.label = tk.Label(parent, text="Give values")
        self.label.grid(row="1")

        self.entries()
        self.labels()

    def entries(self):
        for i in range(5):
            number1 = tk.StringVar()
            entry = tk.Entry(textvariable=number1)
            entry.grid(row=3, column=i)

            result = entry.get()
            return result

    def labels(self,):
        for i in range(5):
            label = tk.Label(self.parent, text=self.entries(), width=17, borderwidth=2, relief="groove")
            label.grid(row=4, column=i)

if __name__ == "__main__":
    root = tk.Tk()
    root.geometry("1280x800")
    MainApplication(root).grid()
    root.mainloop()

The output of my code is following. Apparently, lots of things are wrong because I don't get five entry boxes and they don't update to labels below automatically. enter image description here

I have two questions:

  1. How to fix my code to get wanted output
  2. Would it be a better way to build the GUI using nested classes inside my MainApplication class instead of methods.
Mkk0
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  • Your calling `entries()` when the class is initially called, which means, there are no data inside of the entries yet and they remain empty. I think the best way here is to use, `trace()` for `StringVar()`. And yes there might be a better way to do this. – Delrius Euphoria Nov 28 '20 at 10:11
  • If your sure to have just 5 entry boxes, then just make 5 boxes instead of looping through it. – Delrius Euphoria Nov 28 '20 at 10:22
  • I'm trying to have the program easily scalable and it doesn't feel that if I will manually create a new entry box. – Mkk0 Nov 28 '20 at 10:25
  • If your looping through it, the only solution i can think of is "recursion". – Delrius Euphoria Nov 28 '20 at 10:35

1 Answers1

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If you want labels to have exactly the same text as the entry widgets, the simplest solution by far is to have them share the same textvariable.

Here is a basic example. When you run it and then type into any entry, the value is immediately displayed in the label below it.

import tkinter as tk

class MainApplication(tk.Frame):
    def __init__(self, parent, *args, **kwargs):
        tk.Frame.__init__(self,parent, *args, **kwargs)
        self.parent = parent

        self.label = tk.Label(parent, text="Give values")
        self.label.grid(row=1)

        for column in range(5):
            var = tk.StringVar()
            entry = tk.Entry(root, textvariable=var)
            label = tk.Label(root, textvariable=var, width=17)

            entry.grid(row=2, column=column, sticky="ew")
            label.grid(row=3, column=column, sticky="ew")


if __name__ == "__main__":
    root = tk.Tk()
    root.geometry("1280x800")
    MainApplication(root).grid()
    root.mainloop()

I don't normally advocate for using the special tkinter variables for Entry widgets, but this is a good example of when it is the right tool for the job.

screenshot

As for the question about using nested classes, no, they wouldn't help. There is almost never a time when nested classes are better than defining classes at the global scope. The nesting makes the code more complicated to write and to understand while providing very little extra value.

Nested classes can be useful in very specific instances, but shouldn't normally be used.

Bryan Oakley
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