有条件地显示单个TastyPie资源中的全部或部分字段

Question

Can I use a single TastyPie resource and conditionally have it return all or a subset of columns?

I have an employee database, which I can pull a full record via: /api/v1/employee/ . But certain data in this table can change over time (eg, someone moves to a different group, or their job title changes).

We want to store certain data for historical purposes and be able to query metrics in the future -- for example: "how many 'level 1' employees took this test?". But if Bob has been promoted to 'level 2' since taking the test he would no longer appear in this query, if I simply linked to the employee model.

Can I set up my TastyPie resource to conditionally return a subset of fields, such as (pseudo code follows):

class EmployeeResource(ModelResource):
    # bunch of fields

    class Meta:
        if t = true:
            fields = [ ... ]

... and then access via /api/v1/employee/?t=true (or some other addition to the URL).

Or is it just as effective to just create a completely different resource that can be referenced to return the filtered field set?

Answer 1

1. You can create a different resource, subclassed from EmployeeResource
2. You can put data inside custom dehydrate method:

a

class EmployeeResource(ModelResource):
    def dehydrate(self, bundle):
        t = bundle.request.GET.get('t')
        if t:
            bundle.data['custom_field'] = bundle.obj.custom_field
        return bundle

    class Meta:
        fields = common_fields

Answer 2

Meta.fields is used to specify which database columns are returned, not records of data. It sounds like you are looking to filter out certain records based on some condition (ie query for all persons with level = 1):

* http://django-tastypie.readthedocs.org/en/latest/resources.html#basic-filtering

class EmployeeResource(ModelResource):
    class Meta:
        filtering = {
            "level": ('exact',),
        }

Then just specify the filter as a query parameter: /api/v1/employee/?level=1