How to expose a property (virtual field) on a Django Model as a field in a TastyPie ModelResource

Question

I have a property in a Django Model that I'd like to expose via a TastyPie ModelResource.

My Model is

class UserProfile(models.Model):     _genderChoices = ((u"M", u"Male"), (u"F", u"Female"))      user = Models.OneToOneField(User, editable=False)     gender = models.CharField(max_length=2, choices = _genderChoices)      def _get_full_name(self):         return "%s %s" % (self.user.first_name, self.user.last_name)      fullName = property(_get_full_name) 

My ModelResource is

class UserProfileResource(ModelResource):     class Meta:         queryset = models.UserProfile.objects.all()         authorization = DjangoAuthorization()         fields = ['gender', 'fullName'] 

However all I'm currently getting out of the tastypie api is:

{     gender: 'female',     resource_uri: "/api/v1/userprofile/55/" } 

I have tried playing with the fields property in the ModelResource, but that hasn't helped. Would love to understand what is going on here.

Answer 1

You should be able to define it as a field try:

class UserProfileResource(ModelResource):     fullname = fields.CharField(attribute='_get_full_name', readonly=True)     class Meta:         queryset = models.UserProfile.objects.all()         authorization = DjangoAuthorization()         fields = ['gender',] 

Edit

You also have to include: set readonly=True on your CharField, or TastyPie will try to set its value on insertion or update.

Answer 2

A full example with dehydrate:

class UserResource(ModelResource):     fullname = fields.CharField(readonly=True)      class Meta:         queryset = auth_models.User.objects.all()         resource_name = 'user'      def dehydrate_fullname(self, bundle):         return u"{first_name} {last_name}".format(             first_name=bundle.obj.first_name, last_name=bundle.obj.last_name)