Re: django template loop collections.defaultdict(lambda: collections.defaultdict(list))

On Friday, August 23, 2013 5:58:38 PM UTC+2, Daviddd wrote:

Dear All,

In my view I create the following dictionary of lists from a queryset

#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')

Position field is a charFiled following the format: X-X-X-X used the create the my_dict

my_dict is

--[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

In my template I tried:

{% for key, groups in queryset.iteritems %}      groups = {{ groups }} <br>      {% for group_key, cols in groups.iteritems %}        cols = {{ group_key }} <br>        {% for objs in cols %}            {# rest of the code #}  

But only the first loop is evaluated

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>]}))    groups = (2, defaultdict(<type 'list'>, {1: [<Obj: Obj 7.7 by daviddd>]}))    groups = (3, defaultdict(<type 'list'>, {1: [<Obj: Obj 7.8 by daviddd>]}))  

(Linked question: http://stackoverflow.com/questions/18406126/django-template-loop-collections-defaultdictlambda-collections-defaultdictlis)

I'm not able to spot the error.

Thanks

D

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