Select a valid choice. 2.6 is not one of the available choices

Question

I have a model with a field having choices in it as given below. I am messing with the DecimalField somehow. And i cannot pass values like 1.7 or 2.6. But it is accepting 1.0 and 2.0 . Please let me know what am I doing wrong.

class Trip(models.Model):

    ACE = 1.00
    Dosth = 1.70
    TEN = 2.00
    FORTEEN = 2.30
    SEVENTEEN = 2.60
    NINETEEN = 2.90
    TWENTY = 3.10
    TWENTYTWO = 3.10

    VEH_CHOICES = (
        (ACE, 'ACE'),
        (Dosth, 'Dosth'),
        (TEN, '10.5FT'),
        (FORTEEN, '14FT'),
        (SEVENTEEN, '17FT'),
        (NINETEEN, '19FT'),
        (TWENTY, '20FT'),
        (TWENTYTWO, '22FT'),
    )
    ftype = models.DecimalField(null = True, blank = True, verbose_name = "Vehicle Type", decimal_places = 2, max_digits = 5, choices = VEH_CHOICES)

score 2 · Accepted Answer · answered Mar 06 '20 at 12:18

You need to change field DecimalField to FloatField i give the new model with change

class Trip(models.Model):

    ACE = 1.00
    Dosth = 1.7
    TEN = 2.00
    FORTEEN = 2.30
    SEVENTEEN = 2.60
    NINETEEN = 2.90
    TWENTY = 3.10
    TWENTYTWO = 3.10

    VEH_CHOICES = (
        (ACE, 'ACE'),
        (Dosth, 'Dosth'),
        (TEN, '10.5FT'),
        (FORTEEN, '14FT'),
        (SEVENTEEN, '17FT'),
        (NINETEEN, '19FT'),
        (TWENTY, '20FT'),
        (TWENTYTWO, '22FT'),
    )
    ftype = models.FloatField(null = True, blank = True, verbose_name = "Vehicle Type", choices = VEH_CHOICES)

    def __str__(self):
        return str(self.ftype)

see diff of DecimalField and FloatField

Update this code to your code see it work. if it's work perfect or not let me know

Worked well. Thanks for your time brother :) – Srijwal R Mar 06 '20 at 12:22 — Srijwal R, Mar 06 '20 at 12:22

Select a valid choice. 2.6 is not one of the available choices

1 Answers1