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UUID field added after data already in database. Is there any way to populate the UUID field for existing data?

I've added a UUID field to some of my models and then migrated with South. Any new objects I create have the UUID field populated correctly. However the UUID fields on all my older data is null.

Is there any way to populate UUID data for existing data?

like image 592
blinduck Avatar asked Sep 18 '12 02:09

blinduck


3 Answers

To add UUID values to all existing records first you will need to make sure your model has the UUID filed with blank=True, null=True

Then Run the schemamigration command with south and then open up the resulting migration file. And then Edit your migration file with the following as shown in this post

Quote:

You'll need to import the following import uuid

At the end of the forwards() function add the following def forwards(self, orm):

...
for a in MyModel.objects.all():
    a.uuid = u'' + str(uuid.uuid1().hex)
    a.save()

As stated that will loop through existing instances and add a uuid to it as part of the migration.

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Glyn Jackson Avatar answered Nov 10 '22 08:11

Glyn Jackson


For the following sample class:

from django_extensions.db.fields import UUIDField

def MyClass:
    uuid = UUIDField(editable=False, blank=True)
    name = models.CharField()

If you're using South, create a data migration:

python ./manage.py datamigration <appname> --auto

And then use the following code to update the migration with the specific logic to add a UUID:

from django_extensions.utils import uuid

def forwards(self, orm):
    for item in orm['mypp.myclass'].objects.all():
        if not item.uuid:
            item.uuid = uuid.uuid4() #creates a random GUID
            item.save()


def backwards(self, orm):
    for item in orm['mypp.myclass'].objects.all():
        if item.uuid:
            item.uuid = None
            item.save()

You can create different types of UUIDs, each generated differently. the uuid.py module in Django-extensions has the complete list of the types of UUIDs you can create.

It's important to note that if you run this migration in an environment with a lot of objects, it has the potential to time out (for instance, if using fabric to deploy). An alternative method of filling in already existing fields will be required for production environments.

It's possible to run out of memory while trying to do this to a large number of objects (we found ourselves running out of memory and having the deployment fail with 17,000+ objects).

To get around this, you need to create a custom iterator in your migration (or stick it where it's really useful, and refer to it in your migration). It would look something like this:

def queryset_iterator(queryset, chunksize=1000):
    import gc
    pk = 0
    last_pk = queryset.order_by('-pk')[0].pk
    queryset=queryset.order_by('pk')
    if queryset.count() < 1
        return []
    while pk < last_pk:
        for row in queryset.filter(pk__gt=pk)[:chunksize]:
            pk = row.pk
            yield row
        gc.collect()

And then your migrations would change to look like this:

class Migration(DataMigration):

    def forwards(self, orm):
        for item in queryset_iterator(orm['myapp.myclass'].objects.all()):
            if not item.uuid:
                item.uuid = uuid.uuid1()
                item.save()

    def backwards(self, orm):
        for item in queryset_iterator(orm['myapp.myclass'].objects.all()):
            if item.uuid:
                item.uuid = None
                item.save()
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George Stocker Avatar answered Nov 10 '22 10:11

George Stocker


There is now an excellent, updated answer for Django 1.9 to this exact question in the Django docs.

Saved me a lot of time!

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YPCrumble Avatar answered Nov 10 '22 08:11

YPCrumble