Django Admin linking to related objects

user_636064 0
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My app has users who create pages. In the Page screen of the admin, I'd like to list the User who created the page, and in that list, I'd like the username to have a link that goes to the user page in admin (not the Page).

class PageAdmin(admin.ModelAdmin):
    list_display = ('name', 'user', )
    list_display_links = ('name','user',)
admin.site.register(Page, PageAdmin)

I was hoping that by making it a link in the list_display it would default to link to the actual user object, but it still goes to Page.

I'm sure I'm missing something simple here.

django-admin django
7 Answers

I ended up with a simple helper:

from django.shortcuts import resolve_url
from django.utils.safestring import SafeText
from django.contrib.admin.templatetags.admin_urls import admin_urlname
from django.utils.html import format_html


def model_admin_url(obj: Model, name: str = None) -> str:
    url = resolve_url(admin_urlname(obj._meta, SafeText("change")), obj.pk)
    return format_html('<a href="{}">{}</a>', url, name or str(obj))

Then you can use the helper in your model-admin:

class MyAdmin(admin.ModelAdmin):
    readonly_field = ["my_link"]

    def my_link(self, obj):
        return model_admin_url(obj.my_foreign_key)
user_7958560

I decided to make a simple admin mixin that looks like this (see docstring for usage):

from django.contrib.contenttypes.models import ContentType
from django.utils.html import format_html
from rest_framework.reverse import reverse


class RelatedObjectLinkMixin(object):
    """    
    Generate links to related links. Add this mixin to a Django admin model. Add a 'link_fields' attribute to the admin
    containing a list of related model fields and then add the attribute name with a '_link' suffix to the
    list_display attribute. For Example a Student model with a 'teacher' attribute would have an Admin class like this:

    class StudentAdmin(RelatedObjectLinkMixin, ...):
        link_fields = ['teacher']

        list_display = [
            ...
            'teacher_link'
            ...
        ]
    """

    link_fields = []

    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        if self.link_fields:
            for field_name in self.link_fields:
                func_name = field_name + '_link'
                setattr(self, func_name, self._generate_link_func(field_name))

    def _generate_link_func(self, field_name):
        def _func(obj, *args, **kwargs):
            related_obj = getattr(obj, field_name)
            if related_obj:
                content_type = ContentType.objects.get_for_model(related_obj.__class__)
                url_name = 'admin:%s_%s_change' % (content_type.app_label, content_type.model)
                url = reverse(url_name, args=[related_obj.pk])
                return format_html('<a href="{}" class="changelink">{}</a>', url, str(related_obj))
            else:
                return None
        return _func
user_17925800

If anyone is trying to do this with inline admin, consider a property called show_change_link since Django 1.8.

Your code could then look like this:

class QuestionInline(admin.TabularInline):
    model = Question
    extra = 1
    show_change_link = True


class TestAdmin(admin.ModelAdmin):
    inlines = (QuestionInline,)


admin.site.register(Test, TestAdmin)

This will add a change/update link for each foreign key relationship in the admin's inline section.

user_13845450
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