How Do I Show Django Admin Change List View of for

Question

I'm working on an app with a Model heirarchy of Campaign > Category > Account. Ideally I'd like to users to be able to click on a link in the campaign admin list view and go to a url like "/admin/myapp/campaign/2/accounts/" which will show a django admin view with all the handy ChangeList amenities but which is filtered to show just the accounts in categories in the specified campaign (ie. Account.object.filter(category_campaign_id = 2)). (Note, categories themselves I'm happy to just be "filters" on this accounts list view).

I can't seem to find any reference to a way to mimic this item-click-goes-to-list-of-foriegn-key-children approach that is common in many other frameworks.

Is it possible? Is there a "better" approach in the django paradigm?

thanks for any help!

Answer 1

These are good solutions. I wasn't aware of the auto-filter by url paradigm. Here's another I've discovered which allows you use a custom url scheme:

from consensio.models import Account
from django.contrib import admin        
from django.conf.urls.defaults import patterns, include, url

class AccountAdmin(admin.ModelAdmin):
    campaign_id = 0;

    def campaign_account_list(self, request, campaign_id, extra_context=None):
        '''
        First create your changelist_view wrapper which grabs the extra
        pattern matches
        '''
        self.campaign_id = int(campaign_id)
        return self.changelist_view(request, extra_context)

    def get_urls(self):
        '''
        Add your url patterns to get the foreign key
        '''
        urls = super(AccountAdmin, self).get_urls()
        my_urls = patterns('',
            (r'^bycampaign/(?P<campaign_id>\d+)/$', self.admin_site.admin_view(self.campaign_account_list))
        )
        return my_urls + urls

    def queryset(self, request):
        '''
        Filter the query set based on the additional param if set
        '''
        qs = super(AccountAdmin, self).queryset(request)
        if (self.campaign_id > 0):
            qs = qs.filter(category__campaign__id = self.campaign_id)
        return qs

And plus you'd need to incorporate the URL link into CampaignAdmin's list view...

Answer 2

This was an interesting question so I whipped up a sample app to figure it out.

# models.py
from django.db import models

class Campaign(models.Model):
    name = models.CharField(max_length=20)

    def __unicode__(self):
        return unicode(self.name)

class Category(models.Model):
    campaign = models.ForeignKey(Campaign)
    name = models.CharField(max_length=20)

    def __unicode__(self):
        return unicode(self.name)

class Account(models.Model):
    category = models.ForeignKey(Category)
    name = models.CharField(max_length=20)

    def __unicode__(self):
        return unicode(self.name)

# admin.py
from django.contrib import admin
from models import Campaign, Category, Account

class CampaignAdmin(admin.ModelAdmin):
    list_display = ('name', 'related_accounts', )

    def related_accounts(self, obj):
        from django.core import urlresolvers
        url = urlresolvers.reverse("admin:<yourapp>_account_changelist")
        lookup = u"category__campaign__exact"
        text = u"View Accounts"
        return u"<a href='%s?%s=%d'>%s</a>" % (url, lookup, obj.pk, text)
    related_accounts.allow_tags = True
admin.site.register(Campaign, CampaignAdmin)
admin.site.register(Category)

class AccountAdmin(admin.ModelAdmin):
    list_display = ('category', 'name')
    list_filter = ('category__campaign',)
admin.site.register(Account, AccountAdmin)

You'll need to replace with the name of your app where the Account ModelAdmin lives.

Note: the list_filter on the AccountAdmin is required since Django 1.2.4, Django 1.1.3 and Django 1.3 beta 1, which introduced protection from arbitrary filtering via URL parameter in the admin.

Answer 3

If i understand you correctly, you want to add a custom field (a callable in your ModelAdmin's list_display) to your CampaignAdmin change_list view.

Your custom field would be a link that takes the category.id of each category in your change_list and generates a link to the desired, filtered admin view, which seems to be the account-change_list in your case:

admin/yourproject/account/?category__id__exact=<category.id>

Assuming category is a field on your Campaign-Model you could add the follwoing method to your CampaignAdmin:

def account_link(self, obj):
    return '<a href="/admin/yourproject/account/?category__id__exact=%s">Accounts</a>' % (obj.category.id)

account_link.allow_tags = True

And then you add it to the admin's list_display option:

list_display = ('account_link', ...) 

It depends a bit on your data model though.

If you want to create a permanent, filtered change_list view that suits your needs, you may take a look at this article: http://lincolnloop.com/blog/2011/jan/11/custom-filters-django-admin/

Answer 4

The other solutions don't pay attention to the filters you already have applied. They are part of the query string and I wanted to retain them as well.

First you need to get a reference to the request, you can do that by wrapping changelist_view or queryset as I did:

class AccountAdmin(ModelAdmin):
    model = Account
    list_display = ('pk', 'campaign_changelist')

    # ...

    def queryset(self, request):
        self._get_params = request.GET
        return super(AccountAdmin, self).queryset(request)

    def campaign_changelist(self, obj):
        url = reverse('admin:yourapp_account_changelist')
        querystring = self._get_params.copy()
        querystring['campaign__id__exact'] = obj.campaign.pk
        return u'<a href="{0}?{1}">{2}</a>'.format(
            url, querystring.urlencode(), obj.campaign)
    campaign_changelist.allow_tags = True

And something like that will give you a filter inside the changelist rows. Really helpful. :-)