How do I get the actual object id in a Django admi

问题:

I 've already solved the problem of getting the object id being edited using this code:

class CompanyUserInline(admin.StackedInline):
    """
    Defines tabular rules for editing company users direct in company admin
    """
    model = CompanyUser

    def formfield_for_foreignkey(self, db_field, request, **kwargs):

        if db_field.name == "user":
            users = User.objects.filter( Q(is_superuser=False) )
            query = Q()
            for u in users:
                aux = CompanyUser.objects.filter(user=u)
                if aux.count() == 0:
                    query |= Q(pk=u.id)

            try:
                cpu = CompanyUser.objects.filter(company__id=int(request.path.split('/')[4]))
                for p in cpu:
                    query |= Q(pk=p.user.id)
            except:
                pass

            kwargs["queryset"] = User.objects.filter(query).order_by('username')

        return super(CompanyUserInline, self).formfield_for_foreignkey(db_field, request, **kwargs)

But, the int(request.path.split('/')[4]) is really ugly. I want to know how I get the id from the Django AdminModel. I'm sure it's somewhere inside it, anyone knows?

Thank you in advance! ;D

回答1:

As far as i know it is not possible to access the current instance through the formfield_for_...-methods, because they will only be called for a single field instance!

A better point to hook into this logic where you can access the whole instance/form would be get_form. You can also overwrite a form field's queryset there!

回答2:

After some digging around, we were able to grab the arguments that get passed to the admin view (after being parsed by django admin's urls.py) and use that (self_pub_id) to grab the object:

class PublicationAdmin(admin.ModelAdmin):

    def formfield_for_manytomany(self, db_field, request, **kwargs):
        if db_field.name == "authors":
            #this line below got the proper primary key for our object of interest
            self_pub_id = request.resolver_match.args[0]

            #then we did some stuff you don't care about
            pub = Publication.objects.get(id=self_pub_id)
            kwargs["queryset"] = pub.authors.all()
        return super(PublicationAdmin, self).formfield_for_manytomany(db_field, request, **kwargs)

A more elegant solution is to use the accepted answers recomendation and leverage the get_form ModelAdmin member function. Like so:

class ProfileAdmin(admin.ModelAdmin):
    my_id_for_formfield = None
    def get_form(self, request, obj=None, **kwargs):
        if obj:
            self.my_id_for_formfield = obj.id
        return super(ProfileAdmin, self).get_form(request, obj, **kwargs)

    def formfield_for_foreignkey(self, db_field, request, **kwargs):
        if db_field.name == "person":
            kwargs["queryset"] = Person.objects.filter(profile=self.my_id_for_formfield)
        return super(ProfileAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs)

回答3:

I made it work by rewrite change_view()

class CartAdmin(admin.ModelAdmin):

def change_view(self, request, object_id, form_url='', extra_context=None):
    self.object_id = object_id
    return self.changeform_view(request, object_id, form_url, extra_context)


def formfield_for_foreignkey(self, db_field, request, **kwargs):
    print self.object_id
    return super(CartAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs)

then you can call self.object_id inside formfield_for_foreignkey()

回答4:

I made it work by creating a property() in model.py that returns the ID

models.py:

class MyModel(models.Model):
    myfield = models.CharField(max_length=75)
    ...
    def get_id(self):
        return str(self.id)
    getid = property(get_id)

admin.py:

from myapp.models import MyModel

class MyModelAdmin(admin.ModelAdmin):
    list_display = ['mylink',]
    def mylink(self, object):
        return '<a href="http://mywebsite/'+object.getid+'/">Edit</a>'
    mylink.allow_tags = True

admin.site.register(MyModel, MyModelAdmin)

回答5:

The following code snippet will give you the object id:

request.resolver_match.kwargs['object_id']

Sample usage: (I'm filtering the phone numbers shown, to only show customer's phone numbers)

def formfield_for_foreignkey(self, db_field, request, **kwargs):
    if db_field.name == 'preferred_contact_number':
        kwargs['queryset'] = CustomerPhone.objects.filter(customer__pk=request.resolver_match.kwargs['object_id'])
    return super().formfield_for_foreignkey(db_field, request, **kwargs)

P.S: Found it by debugging and walking through accessible variables.

回答6:

I was dealing with a similar situation, and realized that the id that that I needed from the request, I could from from the model it self since it was a foreign key to that model. So it would be something like:

cpu = CompanyUser.objects.filter(company__id=self.company_id)

or what ever the structure of your model dictates.