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]:

Oh I think I see your issue. These model objects, the ones NOT moving are not going to get their values updated in an ipython shell. You have to requery them to see the updated order values. If you use my code and re-query the Steps each time you'll see the movement.

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.