我正在尝试使用现有数据向前迁移模型。该模型有一个新字段，其约束为 unique=True 和 null=False。 当我这么做的时候
./manage.py schemamigration myapp --auto
South 让我通过询问来指定新字段的默认值：
Specify a one-off value to use for existing columns now
通常我将其设置为“无”，但由于该字段需要是唯一的，我想知道是否可以通过以下方式向 South 传递唯一值：
>>> import uuid; uuid.uuid1().hex[0:35]
这给了我一条错误消息
! Invalid input: invalid syntax
通过命令行迁移时是否可以传递 South 随机唯一默认值，有什么想法吗？
谢谢。
不幸的是，只有
datetime但是，您可以通过将其分为三个迁移来实现相同的效果：
数据迁移教程：http://south.readthedocs.org/en/0.7.6/tutorial/part3.html#data-migrations
在 django 1.7+ 中，您可以执行以下操作。它首先添加没有索引且没有唯一性的字段。然后它分配唯一值（我基于名称并使用您需要创建的 slugify 方法），最后再次更改字段以添加索引和唯一属性。
from django.db import migrations
import re
import django.contrib.postgres.fields
from common.utils import slugify
import django.core.validators
def set_slugs(apps, schema_editor):
categories = apps.get_model("myapp", "Category").objects.all()
for category in categories:
category.slug = slugify(category.name)
category.save()
class Migration(migrations.Migration):
dependencies = [
('myapp', '0034_auto_20150906_1936'),
]
operations = [
migrations.AddField(
model_name='category',
name='slug',
field=models.CharField(max_length=30, validators=[django.core.validators.MinLengthValidator(2), django.core.validators.RegexValidator(re.compile('^[0-9a-z-]+$'), 'Enter a valid slug.', 'invalid')], help_text='Required. 2 to 30 characters and can only contain a-z, 0-9, and the dash (-)', unique=False, db_index=False, null=True),
preserve_default=False,
),
migrations.RunPython(set_slugs),
migrations.AlterField(
model_name='category',
name='slug',
field=models.CharField(help_text='Required. 2 to 30 characters and can only contain a-z, 0-9, and the dash (-)', unique=True, max_length=30, db_index=True, validators=[django.core.validators.MinLengthValidator(2), django.core.validators.RegexValidator(re.compile('^[0-9a-z-]+$'), 'Enter a valid slug.', 'invalid')]),
),
]
这里是 Django 关于迁移唯一字段的官方指南。
Migrations that add unique fields
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Applying a "plain" migration that adds a unique non-nullable field to a table
with existing rows will raise an error because the value used to populate
existing rows is generated only once, thus breaking the unique constraint.
Therefore, the following steps should be taken. In this example, we'll add a
non-nullable :class:`~django.db.models.UUIDField` with a default value. Modify
the respective field according to your needs.
* Add the field on your model with ``default=...`` and ``unique=True``
arguments. In the example, we use ``uuid.uuid4`` for the default.
* Run the :djadmin:`makemigrations` command.
* Edit the created migration file.
The generated migration class should look similar to this::
class Migration(migrations.Migration):
dependencies = [
('myapp', '0003_auto_20150129_1705'),
]
operations = [
migrations.AddField(
model_name='mymodel',
name='uuid',
field=models.UUIDField(max_length=32, unique=True, default=uuid.uuid4),
),
]
You will need to make three changes:
* Add a second :class:`~django.db.migrations.operations.AddField` operation
copied from the generated one and change it to
:class:`~django.db.migrations.operations.AlterField`.
* On the first operation (``AddField``), change ``unique=True`` to
``null=True`` -- this will create the intermediary null field.
* Between the two operations, add a
:class:`~django.db.migrations.operations.RunPython` or
:class:`~django.db.migrations.operations.RunSQL` operation to generate a
unique value (UUID in the example) for each existing row.
The resulting migration should look similar to this::
# -*- coding: utf-8 -*-
from __future__ import unicode_literals
from django.db import migrations, models
import uuid
def gen_uuid(apps, schema_editor):
MyModel = apps.get_model('myapp', 'MyModel')
for row in MyModel.objects.all():
row.uuid = uuid.uuid4()
row.save()
class Migration(migrations.Migration):
dependencies = [
('myapp', '0003_auto_20150129_1705'),
]
operations = [
migrations.AddField(
model_name='mymodel',
name='uuid',
field=models.UUIDField(default=uuid.uuid4, null=True),
),
# omit reverse_code=... if you don't want the migration to be reversible.
migrations.RunPython(gen_uuid, reverse_code=migrations.RunPython.noop),
migrations.AlterField(
model_name='mymodel',
name='uuid',
field=models.UUIDField(default=uuid.uuid4, unique=True),
),
]
* Now you can apply the migration as usual with the :djadmin:`migrate` command.
Note there is a race condition if you allow objects to be created while this
migration is running. Objects created after the ``AddField`` and before
``RunPython`` will have their original ``uuid``’s overwritten.
您可以手动编辑迁移文件：
我需要向某些字段添加随机字符，因此我导入了 random 和 randint
import random
import string
并将默认值更改为
default=random.choice(string.lowercase)
它起作用了。
有办法为 South 的每一行设置唯一值。
class Foo(models.Model):
slug = models.SlugField(unique=True, default='')
....
运行 python manage.py schemamigration --auto foo
# Change add_column to this:
db.add_column(u'account_funnel', 'slug',
self.gf('django.db.models.foo.Foo')(default='',
unique=False,
max_length=50),
keep_default=False)
# right above this add such python code:
foos = orm['foo.Foo'].objects.all()
for foo in foos:
foo.slug = slugify(funnel.name)
foo.save()
# Modify slug as unique field
db.create_unique(u'foo_foo', ['slug'])
ps mark this migration as no_dry_run = True
pss do not forget to import slugify function from django.template.defaultfilters import slugify
null=True, unique=Falsemanage.py makemigrationsmanage.py migrateapp\management\commands\fill_uuid.pyfrom django.core.management.base import BaseCommand
from app.models import Product # Replace "Product" for your real model
import uuid;
class Command(BaseCommand):
def handle(self, *args, **options):
all = Product.objects.all()
for instance in all:
instance.now = uuid.uuid4().hex # Replace "now" with your actual field
instance.save()
manage.py fill_uuid