how can I raise an error if user input is a float - Python

Question

I am trying to write a function that checks whether the inputs are a float (decimal) and if so raises an error and returns a messagebox. below are the first two functions, for reference purposed, which preceed the troublesome validate_float().

def validate_integer():
    """Returns error message if input is not an integer"""
    while True:
        try: 
            i.get() == int or j.get() == int
        except Exception:
            return messagebox.showerror('Input Error', 'Input must be integer!')
        else:
            validate_positive_integer()
            break

def validate_positive_integer():
    """Returns error message if input integer is negative"""
    while True:
        if i.get() <= 1 or j.get() <= 1:
            return messagebox.showerror('Input Error', 'Row number input must be greater than 1')
        else:
            validate_float()
            break

The above functions work as intended when it comes to validate_float(), it calls main_function even if the input is a float (eg. 3.4, 6.7, 9.99999 etc). here it is below

def validate_float():
    """Returns error message if input is float"""
    while True:
        if float(i.get()) or float(j.get()):
            return messagebox.showerror('Input Error', 'Row number input cannot be float')
        else:
            main_function()

I just don't want the user to be able to input a float, I've tried a few different ways to get this function to walk, none of which have been 100% to my desired outcome.

You realize that `float('1')` will return `1.0` without error, right? — Mark Ransom, Mar 22 '22 at 21:48
Does this answer your question? [How do you set a conditional in python based on datatypes?](https://stackoverflow.com/questions/14113187/how-do-you-set-a-conditional-in-python-based-on-datatypes) — MisterMiyagi, Mar 22 '22 at 21:55

gianlucazani · Answer 1 · 2022-03-24T09:26:48.027

1

Have you tried with the native python function isinstance()?

Here's an example:

isinstance(1, float) ➔ false
isinstance(0.0, float) ➔ true

Note that when you use float(x) it is just converting x into a float. It will raise OverflowError if you pass a non-convertible value as input (e.g. "hello"), but this cannot be considered to be a proper check.

[EDIT] This solution is correct only if you can assume you will always check numeric values, or string containing numeric values only. It will raise ValueError otherwise. (thanks to G. Anderson for pointing this out)

edited Mar 24 '22 at 09:26

answered Mar 22 '22 at 21:49

gianlucazani

55
6

Minor correction, your example will raise `ValueError: could not convert string to float: "hello"` not `OverflowError` but otherwise good callout on using `isinstance()` – G. Anderson Mar 22 '22 at 22:10
1

Yes sure, in this case the is instance() method is good only if you assume you will always check numeric values. I'll edit the answer, thank you for pointing this out! – gianlucazani Mar 24 '22 at 09:24

score -1 · Answer 2 · answered Mar 22 '22 at 21:49

Your validate_float() function should be as:

def validate_float():
    """Returns error message if input is float"""
    while True:
        if i.get()== float or j.get()== float:
            return messagebox.showerror('Input Error', 'Row number input cannot be float')
        else:
            main_function()

how can I raise an error if user input is a float - Python

2 Answers2