Getting Django admin url for an object

Question

Before Django 1 0 there was an easy way to get the admin url of an object  and I had written a small filter that I d use like this   lt a href     object admin url           gt       lt  a gt   Basically I was using the url reverse function with the view name being  django contrib admin views main change stage   reverse   django contrib admin views main change stage   args  app label  model name  object id      to get the url   As you might have guessed  I m trying to update to the latest version of Django  and this is one of the obstacles I came across  that method for getting the admin url doesn t work anymore   How can I do this in django 1 0   or 1 1 for that matter  as I m trying to update to the latest version in the svn

User · Answer

I solved this by changing the expression to:

reverse( 'django-admin', args=["%s/%s/%s/" % (app_label, model_name, object_id)] )

This requires/assumes that the root url conf has a name for the "admin" url handler, mainly that name is "django-admin",

i.e. in the root url conf:

url(r'^admin/(.*)', admin.site.root, name='django-admin'),

It seems to be working, but I'm not sure of its cleanness.

User · Answer

I had a similar issue where I would try to call reverse  admin index   and was constantly getting django core urlresolvers NoReverseMatch errors   Turns out I had the old format admin urls in my urls py file   I had this in my urlpatterns    r  admin        admin site root     which gets the admin screens working but is the deprecated way of doing it   I needed to change it to this    r  admin    include admin site urls       Once I did that  all the goodness that was promised in the Reversing Admin URLs docs started working

User · Answer

Essentially the same as Mike Ramirez s answer  but simpler and closer in stylistics to django standard get absolute url method  from django urls import reverse  def get admin url self       return reverse  admin  s  s change     self  meta app label  self  meta model name                      args  self id

User · Answer

There s another way for the later versions  for example in 1 10      load admin urls     lt a href     url opts admin urlname  add      gt Add user lt  a gt   lt a href     url opts admin urlname  delete  user pk     gt Delete this user lt  a gt    Where opts is something like mymodelinstance  meta or MyModelClass  meta  One gotcha is you can t access underscore attributes directly in Django templates  like    myinstance  meta     so you have to pass the opts object in from the view as template context

User · Answer

from django core urlresolvers import reverse def url to edit object obj     url   reverse  admin  s  s change     obj  meta app label   obj  meta model name    args  obj id      return u  lt a href   s  gt Edit  s lt  a gt      url   obj   unicode        This is similar to hansen j s solution except that it uses url namespaces  admin  being the admin s default application namespace

User · Answer

If you are using 1 0  try making a custom templatetag that looks like this   def adminpageurl object  link None       if link is None          link   object     return   lt a href    admin  s  s  d   gt  s lt  a gt               instance  meta app label          instance  meta module name          instance id          link          then just use    adminpageurl my object    in your template  don t forget to load the templatetag first

User · Answer

You can use the URL resolver directly in a template  there s no need to write your own filter  E g      url  admin index         url  admin polls choice add         url  admin polls choice change  choice id        url  admin polls choice changelist      Ref  Documentation

User · Answer

For pre 1 1 django it is simple  for default admin site instance    reverse  admin  s  s change     app label  model name   args  object id

User · Answer

Here s another option  using models   Create a base model  or just add the admin link method to a particular model   class CommonModel models Model       def admin link self           if self pk              return mark safe u  lt a target   blank  href            s  s  s   gt  s lt  a gt      self  meta app label                      self  meta object name lower    self pk  self           else              return mark safe u        class Meta          abstract   True   Inherit from that base model     class User CommonModel           username   models CharField max length 765          password   models CharField max length 192    Use it in a template     user admin link      Or view  user admin link

[django] Getting Django admin url for an object

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