How do i created multiple connected objects at once in Django Rest Framework?

Question

I have following model in my Django Model

class Match(models.Model):

    initiator = models.ForeignKey('CustomUser', related_name='match_initiator', on_delete=models.CASCADE)
    matched_person = models.ForeignKey('CustomUser', related_name='matched_person', on_delete=models.CASCADE)
    created_at = models.DateTimeField(auto_now=True)
    algo_version = models.IntegerField()
    room_no = models.ForeignKey('Room', related_name='room_match', on_delete=models.CASCADE, null=True, blank=True)


class Room(models.Model):

    name = models.CharField(max_length=20, unique=True)
    members = models.ManyToManyField(CustomUser, related_name="joined_rooms")
    created_at = models.DateTimeField(auto_now=True,editable=False)
    is_active = models.BooleanField(default=True)

So basically I have two models, Room and Match with Match having a foreign key to a room. Now I want to create both of these objects at once, but I can not do it as first I have to create Room object get the id of it and assign it to Match object instance and then save it again. Problem with this is that if any of the model save is failed, it breaks the code.

Is there any way I can create both Match and Room object at once?

I have found this reference in Django Documentation: But I am not sure whether Django handle it automatically or do I need to handle it.

Thanks in advance!!!

gdef_ · Accepted Answer · 2019-04-11 19:56:34Z

1

You can use atomic transactions to save both objects or none if something fails.

from django.db import IntegrityError, transaction

def your_function(...):
    try:
        with transaction.atomic():
            room = Room.objects.create(...)
            match = Match.objects.create(room_no=room, ...)
            ....
    except IntegrityError:
        handle_exception()

If something inside the transaction fails nothing will be saved.

edited Apr 11, 2019 at 19:56

answered Apr 11, 2019 at 19:41

gdef_

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1 Comment

hardik24 Over a year ago

Is there any way to do it via serializers?

schillingt · Accepted Answer · 2019-04-11 19:36:04Z

1

You'll need to handle it yourself if your serializers are nested.

What it may look like is the following. This is entirely a guess since I don't have existing serializers to work off of, but I hope it'll help you understand the goal.

class MatchSerializer(serializers.ModelSerializer):
    room = RoomSerializer()

    class Meta:
        model = Match
        fields = (...)

    def create(self, validated_data):
        room_data = validated_data.pop('room')
        room = Room.objects.create(**profile_data)
        validated_data['room'] = room
        return super().create(self, validated_data)

You'll also need to handle the update method which was also mentioned in the documentation you linked to.

answered Apr 11, 2019 at 19:36

schillingt

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4 Comments

hardik24 Over a year ago

Will this be an atomic transaction? I mean if lets say room is created but match object failed to created. Will it be reversible?

schillingt Over a year ago

I believe Django wraps the whole request in a transaction by default.

hardik24 Over a year ago

so what is the advantage of using transaction.atomic() as in @gdef_ answer

schillingt Over a year ago

Looks like I'm wrong. It's defaulted to false. docs.djangoproject.com/en/2.2/ref/settings/… You can read more about Django's transaction management here: docs.djangoproject.com/en/2.2/topics/db/transactions/…

VnC · Accepted Answer · 2019-04-11 19:42:34Z

0

You can simply use:

try:
    room = Room.objects.create(...)
    Match.objects.create(room=room, ...)
except:
    print('Catch exception here...')

answered Apr 11, 2019 at 19:42

VnC

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1 Comment

hardik24 Over a year ago

But will it be atomic? I mean if lets say room is created but match object failed to created, then I want Room object to get destroyed as well

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