Django，如何生成没有模型的管理面板？

I'm building a rather large project, that basically consists of this:

我正在构建一个相当大的项目,基本上由以下内容组成:

Server 1: Ice based services. Glacier2 for session handling. Firewall allowing access to Glacier2.

服务器1:基于冰的服务。 Glacier2用于会话处理。防火墙允许访问Glacier2。

Server 2: Web interface (read, public) for Ice services via Glacier2. Admin interface for Ice services via Glacier 2.

服务器2:通过Glacier2为Ice服务提供的Web界面(读取,公共)。通过Glacier 2提供Ice服务的管理界面。

The point I'm concerned with is the web interface. I want to use Django, because it's both written in python and has that incredibly useful automatic admin panel generator.

我关心的是网络界面。我想使用Django,因为它都是用python编写的,并且具有非常有用的自动管理面板生成器。

The web interface doesn't access any database. It connects to an Ice service on Server #1 via the Glacier2 router and uses the API exposed by those services to manipulate data.

Web界面不访问任何数据库。它通过Glacier2路由器连接到Server#1上的Ice服务,并使用这些服务公开的API来操作数据。

And as you probably know, the admin generation in Django depends on the use of Django's ORM; which I'm not using since I have no database to access.

正如您可能知道的那样,Django中的admin生成依赖于Django的ORM的使用;我没有使用,因为我没有数据库可以访问。

So I need to generate the admin panel, but, instead of having an standard data access like the ORM normally does, I need to intercept any "db-access" calls and transform them into Ice service calls, and then take the service's output (if any), transform it into whatever the ORM normally returns and return control to Django.

所以我需要生成管理面板,但是,我不需要像ORM那样进行标准数据访问,而是需要拦截任何“db-access”调用并将它们转换为Ice服务调用,然后获取服务的输出(如果有的话,将其转换为ORM通常返回的任何内容并将控制权返回给Django。

Anyone knows how I could do this? what would I need to subclass? Any specific ideas?

谁知道我怎么能这样做?我需要什么子类?任何具体的想法?

Thanks for your time.

谢谢你的时间。

4 个解决方案

#1

I think there might be a simpler way than writing custom ORMS to get the admin integration you want. I used it in an app that allows managing Webfaction email accounts via their Control Panel API.

我认为可能有一种比编写自定义ORMS更简单的方法来获得所需的管理集成。我在一个允许通过其控制面板API管理Webfaction电子邮件帐户的应用程序中使用它。

Take a look at models.py, admin.py and urls.py here: django-webfaction

在这里看看models.py,admin.py和urls.py:django-webfaction

To create an entry on the admin index page use a dummy model that has managed=False

要在管理索引页面上创建条目,请使用managed = False的虚拟模型

使用admin注册该模型。

You can then intercept the admin urls and direct them to your own views.

然后,您可以拦截管理员网址并将其引导至您自己的视图。

This makes sense if the add/edit/delete actions the admin provides make sense for your app. Otherwise you are better off overriding the admin index or changelist templates to include your own custom actions

如果管理员提供的添加/编辑/删除操作对您的应用有意义,这是有意义的。否则,最好覆盖管理索引或更改列表模板以包含您自己的自定义操作

#2

The real power of the contrib.admin is django Forms. In essence, the admin tool is basically auto-generating a Form to match a Model with a little bit of urls.py routing thrown in. In the end it would probably just be easier to use django Forms apart from the admin tool.

contrib.admin的真正力量是django Forms。从本质上讲,管理工具基本上是自动生成一个Form,以匹配一个带有一点urls.py路由的模型。最后,除了管理工具之外,使用django Forms可能更容易。

#3

you can "mock" some class so it look like a model but it does proxy to your APIs

你可以“模拟”某个类,所以它看起来像一个模型,但它确实代理了你的API

f.e.

class QuerysetMock(object):
    def all():
        return call_to_your_api()
    [...]


class MetaMock(object):
     def fields():
         return fields_mock_objects..
     verbose_name = ''
     [...]

class ModelMock(object):
    _meta = MetaMock()
    objects = QuerysetMock()

admin.site.register(ModelMock)

This may work.. but you need to do a lot django.model compatible stuff

这可能有用..但你需要做很多django.model兼容的东西

#4

The django ORM has a pluggable backent, which means that you can write a backend for things that aren't RDBMSes. It's probably a rather large task, but a good place to get started is with Malcolm Tredinnick's talk from DjangoCon 2008, Inside the ORM.

django ORM有一个可插拔的后台,这意味着你可以为不是RDBMS的东西编写后端。这可能是一个相当大的任务,但是开始的好地方是Malcolm Tredinnick在DjangoCon 2008,ORM内部的演讲。

Otherwise, you could bypass the ORM altogether, and write the forms manually for the access you need.

否则,您可以完全绕过ORM,并手动编写表单以进行所需的访问。

#1