Python 如何在Django中以编程方式为给定模型生成CREATETABLE SQL语句？
我需要在Django应用程序中以编程方式为给定的非托管模型生成CREATE TABLE语句（Python 如何在Django中以编程方式为给定模型生成CREATETABLE SQL语句？,python,sql,django,migration,Python,Sql,Django,Migration,我需要在Django应用程序中以编程方式为给定的非托管模型生成CREATE TABLE语句（managed=False） 因为我正在处理遗留数据库，所以我不想创建迁移并使用sqlmigrate /manage.py sql命令在这方面很有用，但它已在Django 1.8中删除 你知道有什么替代方案吗？不幸的是，似乎没有简单的方法可以做到这一点，但为了你的运气，我刚刚成功地为你制作了一个工作片段，挖掘django迁移丛林的内部 只是： 将代码保存到get\u sql\u create\u tabl
managed=Falsesqlmigrate/manage.py sql你知道有什么替代方案吗？不幸的是，似乎没有简单的方法可以做到这一点，但为了你的运气，我刚刚成功地为你制作了一个工作片段，挖掘django迁移丛林的内部 只是：
get\u sql\u create\u table.py$export DJANGO\u SETTINGS\u MODULE=yourproject.SETTINGSpython get\u sql\u create\u table.py yourapp.yourmodel启动脚本正如建议的那样，我为这个案例发布了一个完整的答案，这个问题可能暗示了这一点 假设您有一个外部DB表，您决定将其作为Django模型访问，因此将其描述为非托管模型（）。 稍后，您需要能够在代码中创建它，例如使用本地数据库进行一些测试。显然，Django不会对非托管模型进行迁移，因此不会在测试数据库中创建它。 这可以通过使用DjangoAPI来解决，而无需求助于原始SQL-。请参见下面更完整的示例，但作为简短的回答，您可以这样使用它：
from django.db import connections
with connections['db_to_create_a_table_in'].schema_editor() as schema_editor:
schema_editor.create_model(YourUnmanagedModelClass)
# your_app/models/your_model.py
from django.db import models
class IntegrationView(models.Model):
"""A read-only model to access a view in some external DB."""
class Meta:
managed = False
db_table = 'integration_view'
name = models.CharField(
db_column='object_name',
max_length=255,
primaty_key=True,
verbose_name='Object Name',
)
some_value = models.CharField(
db_column='some_object_value',
max_length=255,
blank=True,
null=True,
verbose_name='Some Object Value',
)
# Depending on the situation it might be a good idea to redefine
# some methods as a NOOP as a safety-net.
# Note, that it's not completely safe this way, but might help with some
# silly mistakes in user code
def save(self, *args, **kwargs):
"""Preventing data modification."""
pass
def delete(self, *args, **kwargs):
"""Preventing data deletion."""
pass
# your_app/tests/some_test.py
# This will allow to access the `SchemaEditor` for the DB
from django.db import connections
from django.test import TestCase
from your_app.models.your_model import IntegrationView
class SomeLogicTestCase(TestCase):
"""Tests some logic, that uses `IntegrationView`."""
# Since it is assumed, that the `IntegrationView` is read-only for the
# the case being described it's a good idea to put setup logic in class
# setup fixture, that will run only once for the whole test case
@classmethod
def setUpClass(cls):
"""Prepares `IntegrationView` mock data for the test case."""
# This is the actual part, that will create the table in the DB
# for the unmanaged model (Any model in fact, but managed models will
# have their tables created already by the Django testing framework)
# Note: Here we're able to choose which DB, defined in your settings,
# will be used to create the table
with connections['external_db'].schema_editor() as schema_editor:
schema_editor.create_model(IntegrationView)
# That's all you need, after the execution of this statements
# a DB table for `IntegrationView` will be created in the DB
# defined as `external_db`.
# Now suppose we need to add some mock data...
# Again, if we consider the table to be read-only, the data can be
# defined here, otherwise it's better to do it in `setUp()` method.
# Remember `IntegrationView.save()` is overridden as a NOOP, so simple
# calls to `IntegrationView.save()` or `IntegrationView.objects.create()`
# won't do anything, so we need to "Improvise. Adapt. Overcome."
# One way is to use the `save()` method of the base class,
# but provide the instance of our class
integration_view = IntegrationView(
name='Biggus Dickus',
some_value='Something really important.',
)
super(IntegrationView, integration_view).save(using='external_db')
# Another one is to use the `bulk_create()`, which doesn't use
# `save()` internally, and in fact is a better solution
# if we're creating many records
IntegrationView.objects.using('external_db').bulk_create([
IntegrationView(
name='Sillius Soddus',
some_value='Something important',
),
IntegrationView(
name='Naughtius Maximus',
some_value='Whatever',
),
])
# Don't forget to clean after
@classmethod
def tearDownClass(cls):
with connections['external_db'].schema_editor() as schema_editor:
schema_editor.delete_model(IntegrationView)
def test_some_logic_using_data_from_integration_view(self):
self.assertTrue(IntegrationView.objects.using('external_db').filter(
name='Biggus Dickus',
))
defaultexternal_DB # your_app/tests/base.py
class PreventMigrationsDBRouter:
"""DB router to prevent migrations for specific DBs during tests."""
_NO_MIGRATION_DBS = {'external_db', }
def allow_migrate(self, db, app_label, model_name=None, **hints):
"""Actually disallows migrations for specific DBs."""
return db not in self._NO_MIGRATION_DBS
# settings/test.py
DATABASES = {
'default': {
'ENGINE': 'django.db.backends.oracle',
'NAME': 'db_name',
'USER': 'username',
'HOST': 'localhost',
'PASSWORD': 'password',
'PORT': '1521',
},
# For production here we would have settings to connect to the external DB,
# but for testing purposes we could get by with an SQLite DB
'external_db': {
'ENGINE': 'django.db.backends.sqlite3',
},
}
# Not necessary to use a router in production config, since if the DB
# is unspecified explicitly for some action Django will use the `default` DB
DATABASE_ROUTERS = ['your_app.tests.base.PreventMigrationsDBRouter', ]
希望这个详细的Django用户友好的新示例能够帮助用户节省时间。什么口味的数据库？像MySQL这样的数据库允许您执行“SHOW CREATE TABLE”语句。我想问一下，当SQL语句是非托管模型时，为什么需要“以编程方式生成”它？如果重点是创建一个开发数据库，则从原始数据库创建一个仅模式的SQL转储（或要求dba这样做），并将其包含在源代码中。如果您的最终目标是创建一个表，例如进行一些测试，请看一看在特定时刻强制Django为您创建它的方法，例如测试类设置，不要自己处理原始SQL。@Nikita你真的应该发布一个答案，这样任何人都可以投票给你这个伟大的答案。如果它有效（我没有测试过），那么与我一年前给出的解决方案相比，它是一个非常好的解决方案；-）@费罗，谢谢：）行。它起作用了。当我需要模拟外部系统的只读视图来测试我的模型时，我偶然发现了这个问题。看着你的答案，我明白了，我忘了考虑使用Django迁移框架API。通过对文档进行更多搜索，我使用了我指出的方法。