将models.py拆分成几个文件 [英] Split models.py into several files

问题描述

我正在尝试将我的应用程序的 models.py 拆分为多个文件:

I'm trying to split the models.py of my app into several files:

我的第一个猜测是这样做:

My first guess was do this:

myproject/
    settings.py
    manage.py
    urls.py
    __init__.py
    app1/
        views.py
        __init__.py
        models/
            __init__.py
            model1.py
            model2.py
    app2/
        views.py
        __init__.py
        models/
            __init__.py
            model3.py
            model4.py

这不起作用，然后我发现 this，但在这个解决方案中我仍然有问题，当我运行 python manage.py sqlall app1 我得到类似的东西:

This doesn't work, then i found this, but in this solution i still have a problem, when i run python manage.py sqlall app1 I got something like:

BEGIN;
CREATE TABLE "product_product" (
    "id" serial NOT NULL PRIMARY KEY,
    "store_id" integer NOT NULL
)
;
-- The following references should be added but depend on non-existent tables:
-- ALTER TABLE "product_product" ADD CONSTRAINT "store_id_refs_id_3e117eef" FOREIGN KEY     ("store_id") REFERENCES "store_store" ("id") DEFERRABLE INITIALLY DEFERRED;
CREATE INDEX "product_product_store_id" ON "product_product" ("store_id");
COMMIT;

我对此不太确定，但我很担心应该添加以下引用，但依赖于不存在的表:

I'm not pretty sure about this, but i'm worried aboout the part The following references should be added but depend on non-existent tables:

这是我的model1.py文件:

This is my model1.py file:

from django.db import models

class Store(models.Model):
    class Meta:
        app_label = "store"

这是我的model3.py文件:

This is my model3.py file:

from django.db import models

from store.models import Store

class Product(models.Model):
    store = models.ForeignKey(Store)
    class Meta:
        app_label = "product"

显然有效，但我在 alter table 中得到了评论，如果我尝试这样做，也会发生同样的事情:

And apparently works but i got the comment in alter table and if I try this, same thing happens:

class Product(models.Model):
    store = models.ForeignKey('store.Store')
    class Meta:
        app_label = "product"

那么，我应该手动运行alter for references吗?这可能会给我带来南方的问题吗?

So, should I run the alter for references manually? this may bring me problems with south?

推荐答案

我会做以下事情:

myproject/
    ...
    app1/
        views.py
        __init__.py
        models.py
        submodels/
            __init__.py
            model1.py
            model2.py
    app2/
        views.py
        __init__.py
        models.py
        submodels/
            __init__.py
            model3.py
            model4.py

然后

#myproject/app1/models.py:
    from submodels/model1.py import *
    from submodels/model2.py import *

#myproject/app2/models.py:
    from submodels/model3.py import *
    from submodels/model4.py import *

但是，如果你没有充分的理由，将model1和model2直接放在app1/models.py中，将model3和model4放在app2/models.py中

But, if you don't have a good reason, put model1 and model2 directly in app1/models.py and model3 and model4 in app2/models.py

---第二部分---

这是 app1/submodels/model1.py 文件:

This is app1/submodels/model1.py file:

from django.db import models
class Store(models.Model):
    class Meta:
        app_label = "store"

因此更正您的 model3.py 文件:

Thus correct your model3.py file:

from django.db import models
from app1.models import Store

class Product(models.Model):
    store = models.ForeignKey(Store)
    class Meta:
        app_label = "product"

已编辑，以防有人再次出现这种情况:查看 django-schedule 以获取执行此操作的项目示例.https://github.com/thauber/django-schedule/tree/master/时间表/模型https://github.com/thauber/django-schedule/

Edited, in case this comes up again for someone: Check out django-schedule for an example of a project that does just this. https://github.com/thauber/django-schedule/tree/master/schedule/models https://github.com/thauber/django-schedule/

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