Python 3: Tkinter: How to change Entry.get() into an integer

Question

I am coding a simple program which converts imperial units into metric units. However, when I use an Entry.get() command, I need to convert it to an integer and when I try to do that

I get this error:

Traceback (most recent call last):
  File "C:/Users/Bill/converter.py", line 18, in <module>
    conversion=int(e.get())
ValueError: invalid literal for int() with base 10: ''

I tried running an extremely basic version of what I was doing (which is written below), but that still caused the same error.

import tkinter
root= tkinter.Tk()

e=tkinter.Entry(root)
e.pack()

b=tkinter.Button(root, command= lambda: print(e.get()))
b.pack()

conversion=int(e.get())
conversion= conversion* 1.8 +32
l = tkinter.Label(root, text=conversion)

top.mainloop()

I only recently started coding again, so the answer is probably something really simple that I missed, but thanks for any answers.

You are calling `e.get()` one millisecond after creating the widget. The user doesn't have time to enter any data. — Bryan Oakley, Sep 04 '17 at 14:11

Right leg · Answer 1 · 2017-09-04T13:56:31.960

The problem, and what the error states, is that the empty string '' cannot be converted to an integer.

In fact, a lot of strings cannot be converted to an integer. In your case, int(e.get()) raises an error because the entry is empty, but int('') raises an error. Therefore, you need to validate your input before converting it, so as to process it only when it contains the string representation of an integer.

You can wrap a try-except in a get_value function:

def get_value(entryWidget):
    value = entryWidget.get()
    try:
        return int(value)
    except ValueError:
        return None

Then, instead of setting lambda: print(e.get()) as a callback to your button, pass lambda: print(get_value(e)). If the value could be parsed as an integer, this will print the result of int(e.get()). If it couldn't, this will print None.

Here is the modified version of your code:

import tkinter
root= tkinter.Tk()

def get_value(entryWidget):
    value = entryWidget.get()
    try:
        return int(value)
    except ValueError:
        return None


e = tkinter.Entry(root)
e.pack()

b = tkinter.Button(root, command=lambda: print(e.get()))
b.pack()

conversion = get_value(e)
if conversion is not None:
    conversion *= 1.8 + 32
l = tkinter.Label(root, text=conversion)

top.mainloop()

This, however, is a bit awkward. Since the content of the entry is caught before the main loop, the latter will always be empty.

When dealing with GUIs, you cannot think sequencially as you usually do. You should rather ask your button to update the content of your label when pressed, so as to have it display the result of the conversion:

import tkinter


def get_value(entryWidget):
    value = entryWidget.get()
    try:
        return int(value)
    except ValueError:
        return None

def convert(value):
    if value is None:
        return None
    else:
        return 1.8*value + 32

def set_label_text(label, entry):
    value = convert(get_value(entry))
    if value is None:
        label['text'] = "Enter an integer"
    else:
        label['text'] = value


root = tkinter.Tk()

e = tkinter.Entry(root)    
l = tkinter.Label(root, text="")
b = tkinter.Button(root, text="Convert", command=lambda: set_label_text(l, e))

e.pack()
l.pack()
b.pack()

root.mainloop()

When I change my code with the help from your answer, it still doesn't work and I get the same error message, but it does print none when the value is not an integer. The error points to this line: `conversion=int(e.get())`. How can I fix this? Thanks — Bill R, Sep 04 '17 at 13:39
You need to get rid of this `int(e.get())`. It should be replaced by `get_value(e)`. The problem is that a direct conversion to `int` will raise an error, that will not be caught. Wrapping this conversion as I did in a function will allow to catch the error. — Right leg, Sep 04 '17 at 13:44
@BillR Please check my edit. I gave you a code that is working. I hope this will help you grasp the basics of a GUI. — Right leg, Sep 04 '17 at 14:00

FooBar167 · Answer 2 · 2017-09-04T15:16:10.843

Try this code:

import sys

if sys.version_info.major < 3:  # for Python 2.x
    import Tkinter as tk
else:  # for Python 3.x
    import tkinter as tk

class Application(tk.Frame):
    def __init__(self, master=None):
        tk.Frame.__init__(self, master)
        self.master.title("Conversion application")
        self.master.geometry("300x100")
        self.x = tk.IntVar()
        self.grid()
        self.createWidgets()

    def createWidgets(self):
        self.e = tk.Entry(self, textvariable=self.x)
        self.e.bind('<Return>', self.convert)
        self.e.focus()
        self.e.grid()
        self.b = tk.Button(self, text="Convert", command=self.convert)
        self.b.grid()
        self.l = tk.Label(self)
        self.l.grid()

    def convert(self, event=None):
        try:
            print(self.x.get())
            conversion = self.x.get()
            conversion = conversion * 1.8 + 32
            self.l.config(text=conversion)
        except tk.TclError:
            self.l.config(text="Not an integer")

app = Application()
app.mainloop()

Python 3: Tkinter: How to change Entry.get() into an integer

2 Answers2

Linked