Linking directly to Wagtail modeladmin views

Question

I'm successfully using Wagtail's wagtail.contrib.modeladmin to make a regular Django model editable from the Wagtail admin. I'd like to hyperlink to the modeladmin "create" and "edit" views from my template. What is the URL "name" I can use to reference these views? There appears to be no urls module in wagtail.contrib.admin and no documentation on this.

Here's my directory app's models.py:

from django.db import models

class Organisation(models.Model):
    title = models.CharField(max_length=255)
    logo = models.ImageField(upload_to='logos', blank=True)
    ...

And my project's urls.py:

from django.conf.urls import include, url

urlpatterns = [
    url(r'^admin/', include(wagtailadmin_urls)),
    ...
    url(r'^directory/', include('directory.urls')),
]

Ben Sturmfels Ben Sturmfels · Accepted Answer · 2017-05-23T01:49:05

The URL name will be of the following form, where [action] is one of index, create, edit, delete, inspect or choose_parent:

[app_label]_[model_name]_modeladmin_[action]

For example:

{% url 'directory_organisation_modeladmin_index' %}
{% url 'directory_organisation_modeladmin_create' %}
{% url 'directory_organisation_modeladmin_edit' instance_pk=37 %}
{% url 'directory_organisation_modeladmin_delete' instance_pk=37 %}
{% url 'directory_organisation_modeladmin_inspect' instance_pk=37 %}
{% url 'directory_organisation_modeladmin_choose_parent' %}

Since the URLConf provided doesn't contain a namespace argement to include(), no namespace is required.

Similar to Django's built-in Admin, Wagtail generates these URLs dynamically in the ModelAdmin.get_admin_urls_for_registration() with some helpers like AdminURLHelper.get_action_url_name().

Linking directly to Wagtail modeladmin views

1 Answers