Django + SQLite + ForeignKey（'self'）=失敗遷移

Question

在使用SQLite作爲數據庫後端的Django 1.9中，當嘗試在修改模型以使用多表繼承而不是OneToOneField後應用遷移時出現錯誤它以前使用的關係。

特別是，該問題似乎是由於在模型中包含ForeignKey('self')。

models.py

這裏是從該初始遷移成功建立並應用於該應用：

from django.db import models 
from django.contrib.auth.models import User 


class Customer(models.Model): 
    account = models.OneToOneField(User) 
    parent = models.ForeignKey('self', null=True) 

models.py（經修飾）

該應用程序，然後修改爲繼承從用戶模型而不是鏈接到它：

from django.db import models 
from django.contrib.auth.models import User 

class Customer(User): 
    parent = models.ForeignKey('self', null=True) 

在此點T A ./manage.py makemigrations <app>成功，但然後將遷移失敗：

錯誤

Traceback (most recent call last): 
    File "./manage.py", line 10, in <module> 
    execute_from_command_line(sys.argv) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/core/management/__init__.py", line 353, in execute_from_command_line 
    utility.execute() 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/core/management/__init__.py", line 345, in execute 
    self.fetch_command(subcommand).run_from_argv(self.argv) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/core/management/base.py", line 348, in run_from_argv 
    self.execute(*args, **cmd_options) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/core/management/base.py", line 399, in execute 
    output = self.handle(*args, **options) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/core/management/commands/migrate.py", line 200, in handle 
    executor.migrate(targets, plan, fake=fake, fake_initial=fake_initial) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/db/migrations/executor.py", line 92, in migrate 
    self._migrate_all_forwards(plan, full_plan, fake=fake, fake_initial=fake_initial) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/db/migrations/executor.py", line 121, in _migrate_all_forwards 
    state = self.apply_migration(state, migration, fake=fake, fake_initial=fake_initial) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/db/migrations/executor.py", line 198, in apply_migration 
    state = migration.apply(state, schema_editor) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/db/migrations/migration.py", line 123, in apply 
    operation.database_forwards(self.app_label, schema_editor, old_state, project_state) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/db/migrations/operations/fields.py", line 62, in database_forwards 
    field, 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/db/backends/sqlite3/schema.py", line 221, in add_field 
    self._remake_table(model, create_fields=[field]) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/db/backends/sqlite3/schema.py", line 181, in _remake_table 
    self.create_model(temp_model) 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/db/backends/base/schema.py", line 250, in create_model 
    to_column = field.remote_field.model._meta.get_field(field.remote_field.field_name).column 
    File "/home/piranha/.virtualenvs/Python_3.5-Django_1.9/lib/python3.5/site-packages/django/db/models/options.py", line 582, in get_field 
    raise FieldDoesNotExist('%s has no field named %r' % (self.object_name, field_name)) 
django.core.exceptions.FieldDoesNotExist: Customer has no field named 'id' 

同樣，這個問題似乎只發生使用SQLite和包括當自引用ForeignKey。我認爲這是因爲SQLite不能ALTER列導致Django重建表，但自引用ForeignKey正在導致重建尋找新模型中不再存在的'id'列。

我試過手動調整每種方式的遷移操作，甚至使用apps.get_model()加上migrations.RunPython來使用模型的歷史版本。似乎沒有任何工作。

如何調整以下遷移，以避免錯誤：

遷移

# -*- coding: utf-8 -*- 
# Generated by Django 1.9.2 on 2016-09-07 01:16 
from __future__ import unicode_literals 

from django.conf import settings 
import django.contrib.auth.models 
from django.db import migrations, models 
import django.db.models.deletion 


class Migration(migrations.Migration): 

    dependencies = [ 
     migrations.swappable_dependency(settings.AUTH_USER_MODEL), 
     ('base', '0001_initial'), 
    ] 

    operations = [ 
     migrations.AlterModelOptions(
      name='customer', 
      options={'verbose_name': 'user', 'verbose_name_plural': 'users'}, 
     ), 
     migrations.AlterModelManagers(
      name='customer', 
      managers=[ 
       ('objects', django.contrib.auth.models.UserManager()), 
      ], 
     ), 
     migrations.RemoveField(
      model_name='customer', 
      name='account', 
     ), 
     migrations.RemoveField(
      model_name='customer', 
      name='id', 
     ), 
     migrations.AddField(
      model_name='customer', 
      name='user_ptr', 
      field=models.OneToOneField(auto_created=True, default=None, on_delete=django.db.models.deletion.CASCADE, parent_link=True, primary_key=True, serialize=False, to=settings.AUTH_USER_MODEL), 
      preserve_default=False, 
     ), 
    ] 

而且，是的，我知道我最終會需要打破遷移分成多個階段，並做了數據遷移到新的user_ptr字段中。

Answer 1

好吧，我花了一段時間才能出來：但基本上，我不認爲你可以這樣做：

class Customer(User): 
    parent = models.ForeignKey('self', null=True) 

相反，它應該是更喜歡（未經測試:)

class Customer(User): 
    parent = models.ForeignKey('admin.User', null=True) 

如果你想確保數據的完整性實際上是一個客戶，你可以把一個預先驗證的保存（或東西，如果你正在使用某種形式的拯救流水線）

如果您包括數據遷徙也可以通過使用臨時值將其存儲（或轉儲並重新插入對象）來保留該值。