Functions always returns 2

Question

def counter(message):

    if 'dos' or '2' in message:
        return 2

    elif 'tres' or '3' in message:
        return 3

    elif 'cuatro' or '4' in message:
        return 4

    elif 'cinco' or '5' in message:
        return 5

    else:
        return 1

I've wrote this function for to know if there's any number in the text in order to know how many of the same product does the user want, and for some reason it's only returning 2.

If I write it with each statement in parenthesis ( if ('dos' or '2') in message:) it always returns 1.

It's doing it even when '2' or 'dos' is not in the text. I've even tried '3' or '4' as the input and it still returns 2

Any idea why this is happening?

`'dos' or '2' in message` is the same as `True or ('2' in message)`. Since that first thing is true, the `or` itself is true and this `if` is aways the one used. — tdelaney, Oct 17 '21 at 18:23

score 0 · Accepted Answer · answered Oct 17 '21 at 18:21

0

'dos' is always True --> this is why you get always 2

Change to:

 if 'dos' in message or '2' in message:
        return 2

answered Oct 17 '21 at 18:21

balderman

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No, it's doing it when 'dos' or '2' is not in the text – Gonzalo Martín Oct 17 '21 at 18:23
Read again my answer and the code change I propose. – balderman Oct 17 '21 at 18:24
`bool('dos')` is always True. `dos` is "truthy" when evaluated as a boolean. [This answer](https://stackoverflow.com/a/54621800/15497888) has a fair amount of explanation about this type of thing. – Henry Ecker Oct 17 '21 at 18:25
@GonzaloMartín Glad I managed to help. You can accept the answer. – balderman Oct 17 '21 at 18:29

Functions always returns 2

1 Answers1