장고 템플릿 루프 collections.defaultdict : 내보기에서

Question

가 나는

#view.py 
queryset = MyModel.objects.filter(owner=user, dashboard=tab).order_by('position') 
my_dict = collections.defaultdict(lambda: collections.defaultdict(list))  
for obj in queryset: 
    my_dict[int(obj.position.split('-')[0])][int(obj.position.split('-')[2])].append(obj) 
return Response({'queryset': dict(my_dict)}, template_name='dashboard/_maps_tab.html') 

위치 필드 charFiled의 형식은 다음되는 검색어 세트에서 목록의 다음 사전을 만들 (람다 collections.defaultdict (목록)) 가 my_dict을 만들

my_dict는 내가하고 싶은 내 템플릿에서

--[1] #group 
----[1] #col 1 
------ object1.1.1 #group.col.pk 
------ object1.1.2 
------ object1.1.3 
----[2] #col 2 
------ object1.2.4 
----[3] #col3 
------ object1.3.5 
------ object1.3.6 

--[2] #group 
----[1] #col 1 
------object2.1.7 #group.col.pk 

--[3] #group 
----[1] #col1 
------ object3.1.8 #group.col.pk 
----[2] #col2 
------object3.2.9 
------object3.2.10 

이다

,451,515,

{% for groups in queryset.iteritems %} 

    groups = {{ groups }} <br> 

    {% for cols in groups %} 

     cols = {{ cols }} <br> 

     {% for objs in cols %} 

     {{ objs }} in <br><br> 

     {% for obj in objs %}   
      {{ obj.title }}, 
      {{ obj.desc}}, 
      {{ obj.fieldN }},   

     {% endfor %} 

     {% endfor %} 


    {% endfor %} 

{% endfor %} 

결과는 내가 {% for cols in groups.1 %}을하려고 노력하지만 (빈) 작동하지 않습니다

groups = (1, defaultdict(<type 'list'>, {1: [<Obj: Obj 1 by daviddd>, <Obj: Obj 2 by daviddd>, <Obj: Obj3 by daviddd>], 2: [<Obj: Obj 4 by daviddd>], 3: [<Obj: Obj 5 by daviddd>, <Obj: Obj 6 by daviddd>, <Obj: Obj 7 by daviddd>]})) 
cols = 1 
cols = defaultdict(<type 'list'>, {1: [<Obj: Obj 1 by daviddd>, <Obj: Obj 2 by daviddd>, <Obj: Obj3 by daviddd>], 2: [<Obj: Obj 4 by daviddd>], 3: [<Obj: Obj 5 by daviddd>, <Obj: Obj 6 by daviddd>, <Obj: Obj 7 by daviddd>]}) 

groups = (2, defaultdict(<type 'list'>, {1: [<Obj: Obj 7.7 by daviddd>]})) 
cols = 2 
cols = defaultdict(<type 'list'>, {1: [<Obj: Obj 7.7 by daviddd>]}) 

groups = (3, defaultdict(<type 'list'>, {1: [<Obj: Obj 7.8 by daviddd>]})) 
cols = 3 
cols = defaultdict(<type 'list'>, {1: [<Obj: Obj 7.8 by daviddd>]}) 

입니다. {% in cols for groups.iteritems %} 할 경우 : "Int is iterable"입니다.

https://code.djangoproject.com/ticket/16335 내 경우를 보면이

my_dict = collections.defaultdict(lambda: collections.defaultdict(list))  
dictionary['foo']['foo1'].append('bar') 

내가 어떻게 해결할 수있다?

미리 감사드립니다.

D

Answer 1

내 view.py 솔루션이었다 https://code.djangoproject.com/ticket/16335에 링크 된

my_dict = collections.defaultdict(lambda: collections.defaultdict(list)) 

    for obj in queryset: 
     my_dict[int(obj.position.split('-')[0])][int(obj.position.split('-')[2])].append(obj) 

    for obj in my_dict:    
     my_dict[obj].default_factory = None 

    return Response({'queryset': dict(my_dict)}, template_name='_internal_template.html') 

및 Django template can't loop defaultdict

내 템플릿

{% for groups in queryset.itervalues %} 

    groups = {{ groups }}  
    <br><br> 

    {% for cols in groups.itervalues %} 

     cols = {{ cols }}  
     <br><br> 

     {% for obj in cols %} 

     obj = {{ obj}} in <br><br> 
     obj info = {{ obj.title }}, {{ obj.abstract }}<br> 

     {% endfor %}  

    {% endfor %} 

{% endfor %}