[Solved]-How to link to a different object fom a Django admin view?

by

17👍

✅

Define a custom method on the admin class and reference that in your list_display tuple.

from django.core.urlresolvers import reverse
class MyCommentsAdmin(admin.ModelAdmin):

    list_display = ('id', 'name', 'comment', 'content_type', 'object_link', 'ip_address', 'submit_date', 'is_public', 'is_removed')
    list_select_related = True

    def object_link(self, obj):
        ct = obj.content_type
        url = reverse('admin:%s_%s_change' % (ct.app_label, ct.model), args=(obj.id,)) 
        return '<a href="%s">%s</a>' % (url, obj.id)
    object_link.allow_tags = True

Note I’ve added list_select_related=True, as the object_link method references the content_type model, so it would otherwise cause a whole load of extra queries.

👤Daniel Roseman

3👍

This was a good answer but it has changed significantly.
This answer is more applicable to Django 3.2.3.

from django.urls import reverse
from django.utils.safestring import mark_safe

class MyCommentsAdmin(admin.ModelAdmin):

    list_display = (
        'id', 'name', 'comment', 'content_type', 'object_link', 'ip_address',
        'submit_date', 'is_public', 'is_removed'
    )

    def object_link(self, obj):
        app_label = obj._meta.app_label
        model_label = obj._meta.model_name
        url = reverse(
            f'admin:{app_label}_{model_label}_change', args=(obj.id,)
        ) 
        return mark_safe(f'<a href="{url}">{obj.id}</a>')
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