将models.py分解成多个文件 [英] Split models.py into several files

问题描述

我试图将我的应用程序的 models.py 分割成几个文件：

我的第一个猜测是这样做的：

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

这不起作用，于是我发现这个，但在这个解决方案我仍然有一个问题，当我运行 python manage.py sqlall app1 我有一些像：

  BEGIN; 
 CREATE TABLEproduct_product（
idserial NOT NULL PRIMARY KEY，
store_idinteger NOT NULL 
）
; 
  - 应添加以下引用，但取决于不存在的表：
  -  ALTER TABLEproduct_productADD CONSTRAINTstore_id_refs_id_3e117eefFOREIGN KEY（store_id）参考store_store（id ）DEFERRABLE INITIALLY DEFERRED; 
 CREATE INDEXproduct_product_store_idONproduct_product（store_id）; 
 COMMIT; 
  

我不太清楚这一点，但我担心的部分应添加以下引用，但依赖于不存在的表：

这是我的model1.py文件：

从$ d code $ d $ d 
 
 class Store（models.Model）：
 class Meta：
 app_label =store
  

这是我的model3.py文件：

 从django.db导入模型
 
从store.models import Store 
 
 class Product 
 store = models.ForeignKey（Store）
 class Meta：
 app_label =product
  

并且显然有效，但是我在 alter table 中得到了评论，如果我尝试这样的话，会发生同样的事情：

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

那么我应该手动运行参考的参考这可能会给我带来南方的问题？

解决方案

我甚至不能想象为什么你会这样做。但我会假设你有一个很好的理由。如果我因为某些原因需要这样做，我会执行以下操作：

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

然后

来自子模型/ model1.py的$ m $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

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

---第二部分---

这是app1 / submodels / model1.py文件：

 从django.db导入模型
 class Store（models.Model）：
 class Meta：
 app_label =store
  

因此更正您的model3文件：

 从django.db导入模型
 from app1.models import Store 
 
 class Product（models.Model）：
 store = models.ForeignKey（Store）
 class Meta：
 app_label = product
  

编辑，以防再次出现在某人身上：
查看django -schedule为一个仅仅这样做的项目的例子。
https://github.com/thauber/django-schedule / tree / master / schedule / models
https://github.com/ thauber / django-schedule /

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

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

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;

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:

This is my model1.py file:

from django.db import models

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

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"

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"

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

解决方案

I can't even begin to imagine why you'd want to do this. But I'll assume you've got a good reason. If I needed to do this for some reason, I'd do the following:

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

Then

#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 *

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

---second part---

This is app1/submodels/model1.py file:

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

Thus correct your model3 file:

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

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

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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