Tastypie：無法從相關資源獲取一個字段

Question

問題在於ToManyField返回一組對象或一組url。我如何才能從相關表中獲得一個字段？

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 

我試圖與「行動」的queryset的操縱相加法「僅僅」，但它不工作（「不同」 workes罰款），它返回「行動」的所有領域。也許有同樣的方法'唯一'準確地爲tastypie，但我沒有看到。謝謝。

更新：

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

我解決它以這樣的方式，但我希望得到不重複

Answer 1

正常的QuerySet您應該創建一個屬性，並添加屬性名稱爲字段列表。