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copied!<p><strong>UPDATE:</strong> Since Django 1.8, this is built in.</p> <p>See <a href="https://stackoverflow.com/a/28170958/493272">this answer</a> and <a href="https://docs.djangoproject.com/en/1.9/ref/contrib/admin/#django.contrib.admin.InlineModelAdmin.show_change_link" rel="nofollow noreferrer">the official documentation</a>.</p> <p><strong>OLD ANSWER:</strong></p> <p>At the end I found a simple solution.</p> <p>I create a new template called <code>linked.html</code> that is a copy of <code>tabular.html</code> and I added this code to create the link.</p> <pre><code>{% 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 %} </code></pre> <p>then I created a new model <code>LinkedInline</code> inheriting <code>InlineModelAdmin</code></p> <pre><code>#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() </code></pre> <p>Then when I define a new inline, I have only to use my <code>LinkedInline</code> instead of the normal <code>InlineModelAdmin</code>.</p> <p>I hope it can be useful for other people.</p> <p>Giovanni</p>

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