Keeping Django Models Ordered Example

models.py

from django.db import models, transaction
from django.db.models import F, Max


class StepManager(models.Manager):
    """
    Manager to encapsulate bits of business logic
    """

    def move(self, obj, new_order):
        """ Move an object to a new order position """

        qs = self.get_queryset()

        with transaction.atomic():
            if obj.order > int(new_order):
                qs.filter(
                    task=obj.task, order__lt=obj.order, order__gte=new_order
                ).exclude(pk=obj.pk).update(order=F("order") + 1)
            else:
                qs.filter(
                    task=obj.task, order__lte=new_order, order__gt=obj.order
                ).exclude(pk=obj.pk).update(order=F("order") - 1)

            obj.order = new_order
            obj.save()

    def create(self, **kwargs):
        instance = self.model(**kwargs)

        with transaction.atomic():
            # Get our current max order number
            results = self.filter(task=instance.task).aggregate(Max("order"))

            current_order = results["order__max"]

            if current_order is None:
                current_order = 0

            value = current_order + 1

            instance.order = value
            instance.save()

            return instance


class Task(models.Model):
    name = models.CharField(max_length=100)


class Step(models.Model):
    task = models.ForeignKey(Task, related_name="steps", on_delete=models.CASCADE)
    name = models.CharField(max_length=100)
    order = models.IntegerField(default=1)
    objects = StepManager()

Usage in Ipython

from tasks.models import *

t = Task.objects.create(name="Test1")

s1 = Step.objects.create(task=t, name="Testing1")
s2 = Step.objects.create(task=t, name="Testing2")
s3 = Step.objects.create(task=t, name="Testing3")
s4 = Step.objects.create(task=t, name="Testing4")

# Show the existing numbers/structure
In [9]: for s in Step.objects.all():
   ...:     print(f"pk={s.pk} order={s.order} name={s.name}")
   ...:
pk=1 order=1 name=Testing1
pk=2 order=2 name=Testing2
pk=3 order=3 name=Testing3
pk=4 order=4 name=Testing4

# Move the 4th item to the 2nd position 
In [10]: Step.objects.move(s4, 2)

In [11]: for s in Step.objects.all():
    ...:     print(f"pk={s.pk} order={s.order} name={s.name}")
    ...:
pk=1 order=1 name=Testing1
pk=2 order=3 name=Testing2
pk=3 order=4 name=Testing3
pk=4 order=2 name=Testing4

In [12]:

Thanks, I figured it out. Your code worked great, sorry for doubting you :). Thanks for taking time to help me out! I really appreciate it.