Question

I have a django model Level for a game.

class Level(models.Model):

key = models.CharField(max_length=100)
description = models.CharField(max_length=500)
requiredPoints = models.IntegerField()
badgeurl = models.CharField(max_length=100)
challenge = models.ForeignKey(Challenge)

I now want to query for the highest level with a pointsRequired value smaller than a given value.

If i have:

Level 1: requiredPoints: 200
Level 2: requiredPoints: 800
Level 3: requiredPoints: 2000

When I enter e.g. 900 or 1999 as a query parameter, I want level 2 to be returned, when entering 10000 it should be level 3.

in sql it would look like

select   pointsRequired,
abs(pointsRequired - parameter) as closest
from     the_table
order by closest
limit 1

Any tips? Do I have to use an extra Query-Set? How would it look like

Answer 1

I don’t think your SQL is correct. It should return ‘level 3’ if the parameter is 1999. Based on your description the SQL could be:

SELECT pointsRequired FROM the_table WHERE pointsRequired < parameter ORDER BY pointsRequired DESC LIMIT 1;

Or in Django:

try:
    Level.objects.filter(requiredPoints__lt=parameter).order_by('-requiredPoints')[0]
except IndexError: 
    # Do something