Display foreign key columns as link to detail object in Django admin

Question

As explained in link-in-django-admin-to-foreign-key-object, one can display a ForeignKey field as a link to the admin detail page.

To summarize,

class Foo(Model):
    bar = models.ForeignKey(Bar)

class FooAdmin(ModelAdmin):
    list_display = ('link_to_bar',)
    def link_to_bar(self, obj):
        link = urlresolvers.reverse('admin:app_bar_change', args=[obj.bar_id])
        return u'<a href="%s">%s</a>' % (link, obj.bar) if obj.bar else None
    link_to_bar.allow_tags = True

The question is: can we do it more automatically? For instance, provide to the FooAdmin definition a list of foreign key to display as links to detail page:

class FooAdmin(ModelAdmin):
    ...
    list_foreign_key_links = ('bar',)
    ...

I know that these ModelAdmin classes are generated with metaclass programming. Then, it should be possible. What would be a good start to do so?

Answer 1

I wanted it even more automatic, so I can easily list all fields, possibly excluding some, while displaying foreign keys as links. So, building on the linkify function in the reply of @gtalarico, I defined:

def linkified_list_display(model, exclude = []):
    return [linkify(field_name=f.name) if f.many_to_one else f.name 
            for f in model._meta.fields if f.name not in exclude]

Now I can simply do:

class MyModelAdmin(admin.ModelAdmin):
    list_display = linkified_list_display(
        MyModel, exclude = ['not_to_list_1', 'not_to_list_2'])

Answer 2

This is an improvement using this answer, allowing you to supply an optional alternative short description instead of the supplied field_name being used for the column header. And also an optional label_prop value to display an alternative property value of the model if so desired

EDIT: I've also now updated it to handle ManyToManyFields

from typing import Optional

from django.contrib import admin
from django.contrib.contenttypes.models import ContentType
from django.db.models import ManyToManyField, ForeignKey, Model
from django.urls import reverse
from django.utils.html import format_html

def create_link(linked_obj, app_label: str, label_prop: Optional[str] = None) -> str:
    linked_content_type = ContentType.objects.get_for_model(linked_obj)
    model_name = linked_content_type.model
    view_name = f"admin:{app_label}_{model_name}_change"
    link_url = reverse(view_name, args=[linked_obj.pk])
    return "<a href='%s'>%s</a>" % (link_url, getattr(linked_obj, label_prop) if label_prop else linked_obj)


def linkify(field_name: str, label_prop: Optional[str] = None, short_description: Optional[str] = None, as_html_list: bool = False):
    def _linkify(obj: Model):
        content_type = ContentType.objects.get_for_model(obj)
        app_label = content_type.app_label
        field_type = obj._meta.get_field(field_name)
        items = None

        if isinstance(field_type, ManyToManyField):
            items = list(getattr(obj, field_name).all())
        elif isinstance(field_type, ForeignKey):
            items = [getattr(obj, field_name)]
        else:
            print(f'field_name {field_name} is not ManyToManyField or ForeignKey')

        links = [create_link(itm, app_label, label_prop) for itm in items if itm is not None]

        if len(links) > 1:
            if as_html_list:
                html = "<ul>"
                for link in links:
                    html += f'<li>{link}</li>'
                html += "</ul>"
            else:
                html = ", ".join(links)
        else:
            html = links[0]

        return format_html(html)

    _linkify.short_description = [short_description, field_name.replace("_", " ").capitalize()][short_description is None]

    return _linkify

Example usage:

Models

# models.py
from django.db import models

class Country(models.Model):
    name = models.CharField(max_length=200)
    population = models.IntegerField()

class Career(models.Model):
    name = models.CharField(max_length=200)
    average_salary = models.IntegerField()

class Person(models.Model):
    name = models.CharField(max_length=200)
    age = models.IntegerField()
    country = models.ForeignKey(Country, on_delete=models.CASCADE)
    career = models.ForeignKey(Career, on_delete=models.CASCADE)

Admin

@admin.register(Person)
class PersonAdmin(admin.ModelAdmin):
    list_display = [
        "name",
        "age",
        linkify(
          field_name="country",
          label_prop="name",
          short_description="Country Name"
        ),
        linkify(field_name="career"),
    ]

Result

Given the name of Adam's country is France.

| Name | Age |                     Country Name                         |...|
|------|-----|----------------------------------------------------------|...|
| Adam |  20 | <a href="/admin/app/country/123">France</a>              |...|

It's also worth noting...

...the linkify function can also be used along side the display decorator

@admin.register(Person)
class PersonAdmin(admin.ModelAdmin):
    list_display = [
        "name",
        "age",
        "sortable_country_link",
        linkify(field_name="career"),
    ]

    @admin.display(description='Country Name', ordering='country__name')
    def sortable_country_link(self, obj):
        return linkify(field_name='country', label_prop='name')(obj)

Answer 3

The solution below uses this answer but makes it reusable by all models, avoiding the need to add methods to each admin class.

Example Models

# models.py
from django.db import models

class Country(models.Model):
    name = models.CharField(max_length=200)
    population = models.IntegerField()

class Career(models.Model):
    name = models.CharField(max_length=200)
    average_salary = models.IntegerField()

class Person(models.Model):
    name = models.CharField(max_length=200)
    age = models.IntegerField()
    country = models.ForeignKey(Country, on_delete=models.CASCADE)
    career = models.ForeignKey(Career, on_delete=models.CASCADE)

Example Admin

# admin.py
from django.utils.html import format_html
from django.urls import reverse

from .models import Person


def linkify(field_name):
    """
    Converts a foreign key value into clickable links.
    
    If field_name is 'parent', link text will be str(obj.parent)
    Link will be admin url for the admin url for obj.parent.id:change
    """
    def _linkify(obj):
        linked_obj = getattr(obj, field_name)
        if linked_obj is None:
            return '-'
        app_label = linked_obj._meta.app_label
        model_name = linked_obj._meta.model_name
        view_name = f'admin:{app_label}_{model_name}_change'
        link_url = reverse(view_name, args=[linked_obj.pk])
        return format_html('<a href="{}">{}</a>', link_url, linked_obj)

    _linkify.short_description = field_name  # Sets column name
    return _linkify



@admin.register(Person)
class PersonAdmin(admin.ModelAdmin):
    list_display = [
        "name",
        "age",
        linkify(field_name="country"),
        linkify(field_name="career"),
    ]

---

Results

Given an App named app, and a Person instance Person(name='Adam' age=20) with country and carreer foreign key values with ids 123 and 456, the list result will be:

| Name | Age |                          Country                          |...|
|------|-----|-----------------------------------------------------------|...|
| Adam |  20 | <a href="/admin/app/country/123">Country object(123)</a>  |...|

(Continues)

|...|                          Career                         |
|---|---------------------------------------------------------|
|...| <a href="/admin/app/career/456">Career object(456)</a>  |

Answer 4

A slight respin on the accepted answer. It is not necessarily better, but implements some of the advice in the comments:

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


def linkify(field_name):
    def _linkify(obj):
        content_type = ContentType.objects.get_for_model(obj)
        app_label = content_type.app_label
        linked_obj = getattr(obj, field_name)
        linked_content_type = ContentType.objects.get_for_model(linked_obj)
        model_name = linked_content_type.model
        view_name = f"admin:{app_label}_{model_name}_change"
        link_url = reverse(view_name, args=[linked_obj.pk])
        return format_html('<a href="{}">{}</a>', link_url, linked_obj)

    _linkify.short_description = field_name.replace("_", " ").capitalize()
    return _linkify

Answer 5

A good start would be looking at the source of BaseModelAdmin and ModelAdmin. Try to find out how the ModelAdmin generates the default links. Extend ModelAdmin, add a method to generate links to arbitrary foreign keys and look at how ChangeList generates the change list.

I would also suggest you use format_html to render the links, which makes link_to_bar.allow_tags = True unnecessary:

from django.utils.html import format_html

class FooAdmin(ModelAdmin):
    list_display = ('link_to_bar', )
    def link_to_bar(self, obj):
        link = urlresolvers.reverse('admin:app_bar_change', args=[obj.bar_id])
        return format_html('<a href="{}">{}</a>', link, obj.bar) if obj.bar else None