Does django-fsm call save() method after state changed?

Question

I am using django_fsm to manage state in my model. My model looks like:

from django.db import models,
from django_fsm import FSMField, transition


class MyModel(models.Model):
    STATES = (
        ('pending', _('Pending')),
        ('active', _('Active'))
    )  
    state = FSMField(choices=STATES, default='pending', protected=True) 

    @transition(field=state, source='pending', target='active')
    def change_state(self):
        pass

Should I add self.save() to change_state? Will it be called?

score 5 · Accepted Answer · answered Nov 06 '17 at 09:55

5

If calling change_state() succeeds without raising an exception, the state field will be changed, but not written to the database.

So for making changes into database you need to call obj.save() explicitly

def change_view(request, model_id):
    obj = get_object__or_404(MyModel, pk=model_id)
    obj.change_state()
    obj.save()
    return redirect('/')

answered Nov 06 '17 at 09:55

Satendra

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I wonder why this design decision was taken. If it's safe to just save afterwards, Isn't it better to always save it? Is there any tool or way to save it always? – Shiplu Mokaddim Nov 18 '21 at 22:39
@ShipluMokaddim You can add a receiver to `django_fsm.signals.post_transition` that saves the instance. – Fernando Costa Bertoldi Jul 02 '23 at 11:14

score 2 · Answer 2 · answered Sep 08 '22 at 13:59

You can use a post_transition signal to handle this:

from django_fsm.signals import post_transition

@receiver(post_transition, sender=models.MyModel)
def save_new_workflow_state(sender, instance, name, source, target, **kwargs):
    """ Save the new workflow state following successful django_fsm transition. """

    if source != target:
        instance.save()

This comes from this issue.

Does django-fsm call save() method after state changed?

2 Answers2