How to convert a MultiDict to nested dictionary

Question

I would like to convert a POST from Webob MultiDict to nested dictionary. E.g.

So from a POST of:

\'name=Kyle&phone.number=1234&phone.type=ho         


        

Answer 1

I prefer an explicit way to solve your problem:

1. Divide the members which belong to the same structure (or dict) into a same group with same field name, like

   'name=Kyle&phone1=1234&phone1=home&phone2=5678&phone2=work'
   

2. The order of the fields in the form is guaranteed, so the multidict will be: (('name', 'Kyle'), ('phone1', '1234', 'home'), ('phone2', '5678', 'work'))

3. Then the code will be like:

   def extract(key, values):
       extractor = {
          "name":str,
          "phone":lambda *args:dict(zip(('number', 'type'), args)
       }
       trimed_key = re.match(r"^(\w+)", key).group(1)
       return trimed_key, extractor(trimed_key, *values)
   
   nested_dict = {}
   for i in multidict():
       key, values = i[0], i[1:]
       nested_dict.setdefault(key, [])
       trimed_key, data_wanted = extract(key, values) 
       nested_dict[trimed_key].append(data_wanted)
   
   for key in nested_dict:
       if len(nested_dict[key]) == 1:
          nested_dict[key] = nested_dict[key][0]
   

Answer 2

If you have formencode installed or can install it, checkout out their variabledecode module

Answer 3

I haven't had the time to test it and it's quite restrictive, but hopefully this will work (I'm only posting because it's been a while since you posted the question):

>>> def toList(s):
...     answer = []
...     L = s.split("&")
...     for i in L:
...             answer.append(tuple(i.split('=')))
...     return answer

>>> def toDict(L):
...     answer = {}
...     answer[L[0][0]] = L[0][1]
...     for i in L[1:]:
...             pk,sk = L[i][0].split('.')
...             if pk not in answer:
...                     answer[pk] = []
...             if sk not in answer[pk][-1]:
...                     answer[pk][sk] = L[i][1]
...             else:
...                     answer[pk].append({sk:L[i][1]})

If this is not 100%, it should at least get you on a good start.

Hope this helps