Django的模板循环collections.defaultdict：

Question

在我看来，我创建列表下面字典从查询集

#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') 

位置字段是一个charFiled以下格式（拉姆达collections.defaultdict（名单））在创建my_dict

my_dict是

--[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 

在我的模板，我想这样做

{% 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 %} 

结果是

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>]}) 

我试图做{% for cols in groups.1 %}，但它不工作（空）。 如果我这样做{％for cols in groups.iteritems％}我有：“Int不可迭代”。

看着https://code.djangoproject.com/ticket/16335我的情况是

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

我该如何解决？

在此先感谢！

d

Answer 1

我view.py的解决办法是：

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') 

挂https://code.djangoproject.com/ticket/16335和 Django template can't loop defaultdict

我的模板

{% 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 %}