Make link to foreign key object in admin screen

Question

I have class like this which has the ForeignKey

class MyLog(SafeDeleteModel):
    user = models.ForeignKey(CustomUser,on_delete=models.CASCADE)

then I set user in the list_display of MyLog in admin page.

class MyLogAdmin(admin.ModelAdmin):
    list_display = ["id","user"]
class UserAdmin(admin.ModelAdmin):
    list_display = ["id","username"]

Now I want to make link in user of MyLogAdmin page to UserAdmin,

Is it possible?

I think some framework(such as php symfony) administrator system does this automatically.

However is it possible to do this by django admin?

Answer 1

This is, strangely enough, not a builtin feature. But we can effectively do this ourselves with:

from django.utils.html import format_html


class MyLogAdmin(admin.ModelAdmin):
    list_display = ['id', 'username_link']

    @admin.display(ordering='user__username', description='User')
    def username_link(self, obj):
        return format_html(
            '<a href="{}">{}</a>',
            reverse(
                'admin:appname_customuser_change',
                kwargs={'object_id': obj.user_id},
            ),
            obj.user,
        )

with the appname the name of the app where you defined the CustomUser model (which is not per se the same as the MyLog model).

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Note: It is normally better to make use of the settings.AUTH_USER_MODEL ^[Django-doc] to refer to the user model, than to use the User model ^[Django-doc] directly. For more information you can see the referencing the User model section of the documentation.

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Note: We can boost efficiency by fetching the details of related models in the same query in a ModelAdmin with the .list_select_related attribute ^[Django-doc]. In this specific case this thus looks like:

class MyLogAdmin(admin.ModelAdmin):
    list_select_related = ['user']
    # …