CODEWITHSUNDEEP
djangoadmin
Mateusz Jagiełło
Mateusz Jagiełło

Reputation: 7154

Django - custom admin action

I've got model

class Book(models.Model):
    ...
    accepted = models.BooleanField()

and admin

class BookAdmin(admin.ModelAdmin):
    list_display = ('__unicode__', 'accepted')
    actions = ('change_accepted',)

    def change_accepted(self, request, queryset):
            queryset.update(accepted=True)

This create custom action from scrollable action menu. I've need something like this: http://www.theotherblog.com/wp-content/uploads/2009/06/django_admin.jpg (last column). I mean - action from list_display. Is it possible?

Sorry for my english.

Upvotes: 0

Views: 693

Answers (1)

Bernhard Vallant
Bernhard Vallant

Reputation: 50806

I think you're rather looking for calling a method on the admin class through list_display which should display a link in the current row than for an admin action:

class BookAdmin(admin.ModelAdmin):
    list_display = ('__unicode__', 'accepted', 'my_linK')

    def my_link(self, obj):
        return "<a href="%s">View on site</a>" % obj.get_absolute_url()

Upvotes: 1

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