Django InlineModelAdmin: Show partially an inline model and link to the complete model

Question

I defined several models: Journals, volumes, volume_scanInfo etc.

A journal can have more volumes and a volume can have more scanInfo.

What I want to do is:

• in the admin page of journals I want to have the list of the volumes inline (done)
• connect each volume of the previous list to its admin page where I can show the form to edit the volume and the list of its "scan info" inline.

so I want to have something like:

Journal #1 admin page [name] [publisher] [url] ..... list of volumes inline     [volume 10] [..(other fields)..]   <a href="/link/to/volume/10">Full record</a>     [volume 20] [..(other fields)..]   <a href="/link/to/volume/20">Full record</a> 

Then

Volume #20 admin page [volume number] [..(other fields)...] ...... list of the scan info inline     [scan info 33] [..(other fields)..]   <a href="/link/to/scaninfo/33">Full record</a>     [scan info 44] [..(other fields)..]   <a href="/link/to/scaninfo/44">Full record</a> 

What I tried to do is defining a model method that create the code and trying to use it inside the class that defines "volume inline" in the admin, but it doesn't work.

In other words

the model "Volume" has inside something like:

def selflink(self):     return '<a href="/admin/journaldb/volume/%s/">Full record</a>' % self.vid selflink.allow_tags = True 

and

class VolumeInline(admin.TabularInline):     fields = ['volumenumber', 'selflink']     model = Volume     extra = 1 

But this gives the following error:

Exception Value: 'VolumeInline.fields' refers to field 'selflink' that is missing from the form. 

Any idea?

Thanks, Giovanni

Answer 1

UPDATE: Since Django 1.8, this is built in.

See this answer and the official documentation.

OLD ANSWER:

At the end I found a simple solution.

I create a new template called linked.html that is a copy of tabular.html and I added this code to create the link.

{% if inline_admin_form.original.pk %}           <td class="{{ field.field.name }}">               <a href="/admin/{{ app_label }}/{{ inline_admin_formset.opts.admin_model_path }}/{{ inline_admin_form.original.pk }}/">Full record</a>           </td> {% endif %} 

then I created a new model LinkedInline inheriting InlineModelAdmin

#override of the InlineModelAdmin to support the link in the tabular inline class LinkedInline(admin.options.InlineModelAdmin):     template = "admin/linked.html"     admin_model_path = None      def __init__(self, *args):         super(LinkedInline, self).__init__(*args)         if self.admin_model_path is None:             self.admin_model_path = self.model.__name__.lower() 

Then when I define a new inline, I have only to use my LinkedInline instead of the normal InlineModelAdmin.

I hope it can be useful for other people.

Giovanni