class Material(models.Model): name = models.CharField(max_length=50, blank=False) short_name = models.CharField(max_length=2, blank=False, unique=False, default='Al') def __str__(self): return self.name class Meta: db_table = 'material'
I have created this model. Let suppose I have added a new field called
status = models.IntergerField(default=1)
After that If I run command python manage.py makemigrations, then django will add a new field status in table. Let’s I have 3 rows in the table, then value will be as below:
1. Material1, M1, 1 2. Material2, M2,1 3. Material3, M3, 1
With this migration, I also want to change the status of every row and add new row also like
1. Material1, M1, 1 2. Material2, M2,0 3. Material3, M3, 0 4. Material4, M4, 1
Is there any way to handle data manipulation along the schema migration in django?
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Answer
You can set individual value by using Special Operations such as django.db.migrations.RunPython
.
(ref: https://docs.djangoproject.com/en/3.1/ref/migration-operations/#runpython)
Suppose you have added a field ‘status’ to the model and have already completed making the migration(executing manage.py makemigrations
).
Now, execute manage.py makemigrations your_app --empty
.
Then, edit the auto-created migration file like below.
from django.db import migrations def set_individual_status(apps, schema_editor): Material = apps.get_model("your_app", "Material") material_ids_status_0 = [2, 3] material_ids_status_1 = [1, 4] Material.objects.filter(id__in=material_ids_status_0).update(status=0) Material.objects.filter(id__in=material_ids_status_1).update(status=1) class Migration(migrations.Migration): dependencies = [("your_app", "00xx_auto_20210106_1527")] operations = [ migrations.RunPython(set_individual_status, migrations.RunPython.noop) ]
And then, execute manage.py migrate
.