Django Form ChoiceField range(): 'int' object not iterable
Asked Answered
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2

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from django import forms

class SignUpForm(forms.Form):
    birth_day = forms.ChoiceField(choices=range(1,32))

I'm getting "Caught TypeError while rendering: 'int' object is not iterable". https://docs.djangoproject.com/en/dev/ref/models/fields/#field-choices says the choices argument takes iterables such as a list or tuple.

http://docs.python.org/library/functions.html#range says range() creates a list.

Why am I getting an error?

I tried converting the list to str using map() but received different errors.

Shortsighted answered 6/11, 2011 at 5:35 Comment(0)
A
18

... says the choices argument takes iterables such as a list or tuple.

No, it says it takes an iterable of 2-tuples.

An iterable (e.g., a list or tuple) of 2-tuples to use as choices for this field.

birth_day = forms.ChoiceField(choices=((str(x), x) for x in range(1,32)))
Arlenearles answered 6/11, 2011 at 5:38 Comment(3)
Works, looks like I need to be careful when reading. Solution has one extra left parentheses that needs to be removed.Shortsighted
@IgnacioVazquez-Abrams: is there any reason why you use generator instead list or tuple? #16940793Stylopodium
@noisy: Mostly force of habit, but there's no need for the full data structure in the bytecode.Arlenearles
B
0

You need 2 tuples. Use zip built-in function for the same 2 tuples

from django import forms


class SignUpForm(forms.Form):

    birth_day = models.IntegerField(choices=list(zip(range(1, 32), range(1, 32))), unique=True)

Remember (1,32) will create from 1 to 31 !

Bambara answered 17/11, 2017 at 6:43 Comment(0)

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