迁移时传递South随机唯一默认值

5 回答

• 0

  不幸的是，只有 datetime 模块可用作模式移植中的一次性值 .

  但是，通过将其拆分为三个迁移，您可以实现相同的效果：

  • 在没有约束的情况下向模型添加新字段（null = True，unique = False）

  • 使用数据迁移将UUID添加到新字段

  • 在新字段上添加约束（null = False，unique = True）

  数据迁移教程：http://south.readthedocs.org/en/0.7.6/tutorial/part3.html#data-migrations

  回复于 2024-05-07T02:20:30+08:00
• 2

  在django 1.7中，您可以执行以下操作 . 它首先添加没有索引的字段，也没有唯一的 . 然后它分配唯一值（我根据名称和你需要创建的slugify方法），最后再次改变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')]),
          ),
      ]
  

  回复于 2024-05-07T02:20:30+08:00
• 27

  Here是Django关于迁移独特领域的官方方法 .

  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.
  

  回复于 2024-05-07T02:20:30+08:00
• 4

  您可以手动编辑迁移文件：

  我需要在某些字段中添加随机字符，因此我导入了随机和randint

  import random
  import string
  

  并将默认值更改为

  default=random.choice(string.lowercase)
  

  有效 .

  回复于 2024-05-07T02:20:30+08:00
• 0

  有办法为South的每一行做独特的 Value .

  将models.py中的slug定义为：

  class Foo(models.Model):
    slug = models.SlugField(unique=True, default='')
    ....
  

  创建新迁移

  运行python manage.py schemamigration --auto foo

  打开新的迁移文件，然后进行编辑：

  # 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
  

  回复于 2024-05-07T02:20:30+08:00