is it possible to create a custom admin view witho

问题:

I have an object which I want to use under admin instead of a model which inherits models.Model. If I make it inherit models.Model, this object will create a table in the database which i don't want. I only want this object to stay in memory.

One solution I have come with help from the nice people at stack overflow is I create admin views, register these custom views via a modelAdmin ( admin.site.register() ) under admin.py and use this model-like object as dynamic data storage (in memory).

Since this model like object doesn't inherit from models.Model, admin.site.register() (under admin.py) doesnt accept it and shows a 'type' object is not iterable" error when I try to access it in the browser.

回答1:

hmmm. Thanks for your help everyone. The solution I have come up ( with your help ofcourse :) is as follows:

I have two custom templates:

   my_model_list.html
   my_model_detail.html

Under views.py:

class MyModel(object):
    # ... Access other models
    # ... process / normalise data 
    # ... store data

@staff_member_required
def my_model_list_view(request) #show list of all objects
    #. . . create objects of MyModel . . .
    #. . . call their processing methods . . .
    #. . . store in context variable . . . 
    r = render_to_response('admin/myapp/my_model_list.html', context, RequestContext(request))
    return HttpResponse(r)

@staff_member_required
def my_model_detail_view(request, row_id) # Shows one row (all values in the object) in detail     
    #. . . create object of MyModel . . .
    #. . . call it's methods . . .
    #. . . store in context variable . . . 
    r = render_to_response('admin/myapp/my_model_detail.html', context, RequestContext(request))
    return HttpResponse(r)

Under the main django urls.py:

urlpatterns = patterns( 
    '',
    (r'^admin/myapp/mymodel/$', my_model_list_view),
    (r'^admin/myapp/mymodel/(\d+)/$', my_model_detail_view),
    ( r'^admin/', include( admin.site.urls ) )
)

回答2:

You can add your views directly to the AdminSite object, rather than to any particular ModelAdmin subclass which you then register.

The default AdminSite is accessed via django.contrib.admin.site, which is what you call register and autodiscover on. Instead of using this, you could create your own subclass and add your own views to it, and then register your models against that rather than the default one.

回答3:

The most straightforward answer is "no". As the Django Book says, the admin is for "Trusted users editing structured content," in this case the structured content being models arranged in hierarchies and configured through settings.py. More importantly, if your object doesn't completely duck-type to a models.Model complete with expected relationships, the admin will probably toss exceptions all over the place.

However, as the mantra goes, "It's just python." You can override any of the pages in admin. Just create your own templates in your project, and have them come first in the template search. Also, by inheriting admin/base.html, you maintain the look & feel of the administration project.

Write your administrative view and templates for this object, just like any others, but making sure to wrap the views in the is_staff decorator to ensure that the views are protected from access by unauthorized users. Put those in the application, perhaps in admin/views.py, with templates/admin/object_list.html and object_form.html.

Once you have appropriate administrative tools for these non-database objects, you can then provide access to them through the administration index page: You want to override admin/index.html, and provide additional project-specific items to the page as needed.

I have done exactly this to provide administrative access to third-party APIs that store our data, such as the ConstantContact email service, and it works pretty well.