Here is a one-liner example for searching:

aaa = {
  "success":True,
  "data":
  {
    "array":
    [
      {
        "id":"1","name":"Value1"
      },
      {
        "id":"2","name":"Value2"
      }
    ]
  }
}

[a['name'] for a in aaa['data']['array'] if a['id']=='1']

This will return all the found cases, or an empty array if nothing is found

Here’s an easy way to do it:

def function(json_object, name):
    for dict in json_object:
        if dict['name'] == name:
            return dict['price']

If you are sure that there are no duplicate names, an even more effective (and pythonic) way to do it is to use list comprehensions:

def function(json_object, name):
        return [obj for obj in json_object if obj['name']==name][0]['price']

The error message is correct.

key = json.loads(response['password'])
print(key[0]),

The format of json is string. You need to convert the string of a json object to python dict before you can access it.

i.e.: loads(string) before info[key]

key = json.loads(response)['password']
print(key[0])

w = y['filters']['filterB'] doesn't work because y['filters'] is a list and not dict.

The answer to your question depends on how you want to handle the case of multiple dictionaries inside filters list that have filterB key.

import json
x = '{"filters":[{"filterA":"All"},{"filterB":"val1"}]}'
y = json.loads(x)

# all filterB values
filter_b_values = [x['filterB'] for x in y['filters'] if 'filterB' in x.keys()] 

# take first filterB value or None if no values
w = filter_b_values[0] if filter_b_values else None

You have to iterate over the list of dictionaries and search for the one with the given id_number. Once you find it you can print the rest of its data and break, assuming id_number is unique.

data = [
 {
   "id_number": "SA4784",
   "name": "Mark",
   "birthdate": None
 },
 {
   "id_number": "V410Z8",
   "name": "Vincent",
   "birthdate": "15/02/1989"
 },
 {
   "id_number": "CZ1094",
   "name": "Paul",
   "birthdate": "27/09/1994"
 }
]

for i in data:
    if i['id_number'] == 'V410Z8':
        print(i['birthdate'])
        print(i['name'])
        break

If you have control over the data structure, a more efficient way would be to use the id_number as a key (again, assuming id_number is unique):

data =  { "SA4784" : {"name": "Mark", "birthdate": None},
          "V410Z8" : { "name": "Vincent", "birthdate": "15/02/1989"},
          "CZ1094" : {"name": "Paul", "birthdate": "27/09/1994"}
        }

Then all you need to do is try to access it directly:

try:
    print(data["V410Z8"]["name"])
except KeyError:
    print("ID doesn't exist")
>> "Vincent"

You can use a generator expression, calling next on it to supply the first matching item if any:

match = next(d for d in data if d['last_name'] == 'Doe')

json_arr = [
  {
   'Clothes': [
               {
                 'id': '32705111',
                 'describes': 'no problem'
               }
             ]
  },
  {
   'Dress': [
               {
                 'id': '32705111',
                 'describes': 'no outfit'
               }
           ]
  }
]

for d in json_arr:
    for name, array in d.items():
        globals()[name] = array

You can do this:

data =   {
      "BuildId":"4c53575f-36f48a7f1f37",
      "EventId":1,
      "ArchiveLocations":
      [
         {
            "archive_server":"CA",
            "ftp":"ftp://ABC1.com/2011_V2/2011.4.103.002",
            "dfs":"\\\\ABD\\dfs\\Department\\2011.4.103.002"
         },
         {
            "archive_server":"CT",
            "ftp":"ftp://ABC2.com/2011_V2/2011.4.103.002",
            "dfs":"\\\\ABD.com\\dfs\\Department\\2011.4.103.002"
         }
      ]

   }

buildid = data['BuildId']
firstserver = data['ArchiveLocations'][0]['archive_server']
secondserver = data['ArchiveLocations'][1]['archive_server']

print buildid, secondserver

This returns 4c53575f-36f48a7f1f37 CT

Alternately you can iterate over each key and pull the values that way. Its difficult to give you a complete answer because I don't know the context of your question.