Tastypie：无法从相关资源获取一个字段

Question

问题在于ToManyField返回一组对象或一组url。我如何才能从相关表中获得一个字段？

class PlaceResource(ModelResource): 
    location = fields.ManyToManyField(PlaceLocationResource, 'location') 
    action = fields.ToManyField('menus.resources.ActionResource', 
          attribute= lambda bundle: bundle.obj.action.distinct('type').only('name'), 
           null=True, related_name='place', full=True) 
    class Meta: 
     queryset = PlaceInfo.objects.all() 
     resource_name = 'place' 

     filtering = { 
      'id' : ALL, 
     } 

    def dehydrate(self, bundle): 
     #raise sdf 
     bundle.data['type'] = bundle.obj.type.name 
     bundle.data['name'] = bundle.obj.name.name 
     bundle.data['close_time'] = bundle.obj.close_time 
     bundle.data['location'] = {} 
     bundle.data['location']['address'] = (bundle.obj.location.all()[0]).address 
     bundle.data['location']['latitude'] = (bundle.obj.location.all()[0]).latitude 
     bundle.data['location']['longtitude'] = (bundle.obj.location.all()[0]).longtitude 

     if bundle.obj.comments == None: 
      bundle.data['comments'] = 0 
     if bundle.obj.price == None: 
      bundle.data['price'] = 0 
     #if bundle.obj.rate_amount == None: 
     # bundle.data['rate_amount'] = 0 
     #if bundle.obj.rate_makr == None: 
     # bundle.data['rate_makr'] = 0 
     bundle.data['rate_amount'] = { 
      1 : 10, 
      2 : 15, 
      3 : 40, 
      4 : 35, 
      5 : 27 
     } 
     if bundle.obj.open_time == None: 
      bundle.data['open_time'] = 'Unsetted' 
     if bundle.obj.close_time == None: 
      bundle.data['close_time'] = 'Unsetted' 
     if bundle.obj.location == None: 
      bundle.data['location'] = 'Unsetted' 
     if bundle.obj.type == 'restaurant.PlaceType.None': 
      bundle.data['type'] = 'Unsetted' 

     #bundle.obj.action = bundle.obj.action.distinct('type').only('type') 
     #bundle.data['types'] = bundle.obj.action 

     return bundle 

我试图与“行动”的queryset的操纵相加法“仅仅”，但它不工作（“不同” workes罚款），它返回“行动”的所有领域。也许有同样的方法'唯一'准确地为tastypie，但我没有看到。谢谢。

更新：

bundle.data['types'] = [] 
for i in bundle.obj.action.distinct('type'): 
    bundle.data['types'].append(i.type) 

我解决它以这样的方式，但我希望得到不重复

Answer 1

正常的QuerySet您应该创建一个属性，并添加属性名称为字段列表。