无法使用 Django 3.0.3 中的迁移 API 使用 ModelState 和 ProjectState 进行迁移

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问题描述

我正在使用 ProjectState 迁移到表的新属性.我正在尝试了解在 Django 3.0.3 中使用迁移 API 的 ModelState 和 ProjectState.

I am using ProjectState to migrate to a new attributes of a table. I am trying to understand the ModelState and ProjectState using of migrations API in Django 3.0.3.

我无法迁移到具有新字段的新州.有人可以帮助我了解 ProjectState 和 ModelState 使用什么来申请新的 model_definition 迁移工作吗?以下代码不会迁移到数据库，但不会出现任何错误.

I am unable to migrate to the new state which has new fields. Can someone help me with the ProjectState and ModelState usage of what to apply for new model_definition migration to work? The following code does not migrate to DB but doesnt give any error.

我想从一个数据库表状态迁移到另一个状态，并且有一些元数据_meta.

I want to migrate from a DB table state to another state and there are some metadata _meta.

当前数据库状态model_state.fields为:

[('id', <django.db.models.fields.AutoField>)]

添加 fields_attrs 迁移后的未来数据库状态 model_state.fields 应该是使用 models_definition 的:

The future DB state model_state.fields after adding fields_attrs migrations should be this using the models_definition:

[('id', <django.db.models.fields.AutoField>), ('name', <django.db.models.fields.CharField>)]

模型定义代码为:

model_config 对象是

model_config object is

{
 '__module__': 'testmodule', 'app_label': 'testmodule', 
 '__unicode__': <function ModelScript.model_create_config.<locals>.<lambda> at 0x00000221B6FBEF70>, 
 'attrs': {'name': <django.db.models.fields.CharField>}
}

model_definition 是:

model_definition is:

model_definition = type(
                model_item.table_name,
                # TODO: Put this into Database
                # model_config.get("extends"),
                bases,
                model_config
            )

这是我正在使用的代码:

This is the code I am using:

from django.db.migrations.state import ProjectState
from django.db.migrations.migration import Migration
from django.db.migrations.state import ModelState
from django.db.migrations import operations

# model_definition is coming from a function as the following object
model_definition = {'__module__': 'testmodule', 'app_label': 'testmodule', '__unicode__': <function ModelScript.model_create_config.<locals>.<lambda> at 0x000002047275FF70>, 'attrs': {'name': <django.db.models.fields.CharField>}, '__doc__': 'SampleModel(id)', '_meta': <Options for SampleModel>, 'DoesNotExist': <class 'testmodule.SampleModel.DoesNotExist'>, 'MultipleObjectsReturned': <class 'testmodule.SampleModel.MultipleObjectsReturned'>, 'id': <django.db.models.query_utils.DeferredAttribute object at 0x00000204727F9430>, 'objects': <django.db.models.manager.ManagerDescriptor object at 0x00000204727F9490>}

model_state = ModelState.from_model(model_definition)

# field_attrs are all the new fields to be migrated         
for k,v in field_attrs.items():
    model_state.fields.append((k, v))

# Create a fake migration with the CreateModel operation
cm = operations.CreateModel(name=model_state.name, fields=model_state.fields)

migration = Migration("fake_migration", model_state.app_label)
migration.operations.append(cm)

# SHOULD ProjectState be used for the new definition to be APPLIED to DB and HOW?
state = ProjectState()
with db_conn.schema_editor(collect_sql=True, atomic=migration.atomic) as schema_editor:
     # Following create_model also doesnot migrate to Mysql DB
     # Gives a Table exists Error even with root user of mysql
     # schema_editor.create_model(model_definition)

     # Following doesnot migrate to the new required state
     state = migration.apply(state, schema_editor, collect_sql=True)
     # Following gives atomic transaction error if used along with atomic
     # following commit commented gives no error but doesnt migrate
     # db_conn.commit()

我已阅读此内容并使用如何在 Django 中以编程方式为给定模型生成 CREATE TABLE SQL 语句?

I have read this and using How to programmatically generate the CREATE TABLE SQL statement for a given model in Django?

欢迎任何帮助或资源.

更新:我确实尝试了 Django 的测试用例，但它不能以编程方式工作.我必须明确使用 addfield 吗?不确定如何让这个工作.projectstate 和 model_create 方式都不起作用

Update: I did try the test cases of Django and it didn't work programmatically. Do I have to use addfield categorically? Unsure of how to get this working. Both projectstate and model_create way is not working

推荐答案

首先，您需要使用模型元类，即.ModelBase，而不是 type:

To start, you need to be using the model metaclass, ie. ModelBase, and not type:

from django.db.models.base import ModelBase

model_definition = ModelBase(
    model_item.table_name,
    bases,
    model_config
)

一旦你使用了正确的metaclass，你可能会收到无数的错误，因为你使用了很多ModelBase设置的类属性内部，并不指望你设置自己.

Once you use the proper metaclass, you will likely receive a myriad of errors, since you are using many class attributes that ModelBase sets internally, and is not expecting you to set yourself.

您应该只设置 ModelBase 期望在传统模型上设置的属性，而不是转储模型具有的所有属性，其中包括:

Instead of dumping all the attributes that your model has, you should only set the attributes that ModelBase expects to be set on a traditional model, which includes:

__module__ 和 __qualname__
模型字段
自定义管理器或查询集
模型方法
模型元

其他的都应该省略.

例如，如果您有一个看起来像这样的模型，在模块 myapp.models 中:

So for example, if you have a models that look like this, in the module myapp.models:

class Parent(models.Model):
    name = models.CharField(max_length=45)

class Child(models.Model):
    name = models.CharField(max_length=45)
    parent = models.ForeignKey(Parent, on_delete=models.CASCADE)

class ModelWithMeta(models.Model):
    class Meta:
        db_table = 'some_table'

这些模型的动态版本需要如下所示:

The dynamic version of these models need to look like this:

from django.db import models
from django.db.models.base import ModelBase

bases = (models.Model,)

Parent = ModelBase('Parent', bases, {
    '__module__': 'myapp.models',
    '__qualname__': 'Parent',
    'name': models.CharField(max_length=45),
})

Child = ModelBase('Child', bases, {
    '__module__': 'myapp.models',
    '__qualname__': 'Child',
    'name': models.CharField(max_length=45),
    'parent': models.ForeignKey('myapp.Parent', on_delete=models.CASCADE),
})

ModelWithMeta = ModelBase('ModelWithMeta', bases, {
    '__module__': 'myapp.models',
    '__qualname__': 'ModelWithMeta',
    'Meta': type('Meta', (), {'db_table': 'some_table'}),
})

我不明白你的迁移代码的目的，所以我会假设这是一个试图让动态模型工作的黑客，这意味着你可以完全扔掉它并使用内置的迁移加载器，即:

I don't understand the purpose of your migration code, so I will assume that it was a hack in attempt to get the dynamic models working, which means you can probably throw it out altogether and use the builtin migration loader, ie:

python3 manage.py makemigrations myapp && python3 manage.py migrate myapp

我你不熟悉 python 元类，我建议阅读它们，因为这是理解我的代码的先决条件.

I you aren't familiar with python metaclasses, I'd recommend reading up on them, since it's a prerequisite to understand my code.

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无法使用 Django 3.0.3 中的迁移 API 使用 ModelState 和 ProjectState 进行迁移

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