CODEWITHSUNDEEP
pythondjango
Muhammad Tahir
Muhammad Tahir

Reputation: 5174

Link ForeignKey record to its detail view in Django Admin?

If there are two models

models.py

class ModelA(models.Model):
    pass

class ModelB(models.Model):
    a = models.ForeignKey(ModelA, null=True)

admin.py

class ModelBAdmin(admin.ModelAdmin):
    list_display = ('id', 'a', )
    list_display_links = None
    ordering = ('-id', )

admin.site.register(ModelB, ModelBAdmin)


class ModelAAdmin(admin.ModelAdmin):
    list_display = ('id', )
    list_display_links = None

admin.site.register(ModelA, ModelAAdmin)

Is there a way to link a with its detail view in list view of ModelB? If not, then can we do the same in detail view of ModelB? That is, display a as a link to its detail view?

Currently in list view of ModelB there is only id of ModelA. And to view its detail I have to go to ModelA list view and search for the id. What I want id in list view of ModelB to point to detail view of a (like /admin/ModelA/{a.id})

Upvotes: 3

Views: 1076

Answers (1)

evergreen
evergreen

Reputation: 886

The fields in the list view can be callables, so you can do this:

from django.core.urlresolvers import reverse

class ModelBAdmin(admin.ModelAdmin):
    list_display = ('id', 'a_link', )
    list_display_links = None
    ordering = ('-id', )

    def a_link(obj):
        return format_html('<a href="{}">{}</a>', reverse(
            'admin:myapp_a_change', args=[obj.a.id]), obj.a.name)
    a_link.short_description = "Model A Details"

admin.site.register(ModelB, ModelBAdmin)

Upvotes: 2

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