Python decodes JSON

Question

I've the following json:

{
 "slate" : {
 "id" : {
 "type" : "integer"
 },
 "name" : {
 "type" : "string"
 },
 "code" : {
 "type" : "integer",
 "fk" : "banned.id"
 }
 },
 "banned" : {
 "id" : {
 "type" : "integer"
 },
 "domain" : {
 "type" : "string"
 }
 }
}

I'd like to figure out the best decoding way to have an easily browsable python object presentation of it.

I tried:

import json
jstr = #### my json code above #### 
obj = json.JSONDecoder().decode(jstr)
for o in obj:
 for t in o: 
 print (o)

But I get:

 f 
 s
 l
 a
 t
 e
 b
 a
 n
 n
 e
 d

And I don't understand what's the deal. The ideal would be a tree (even a list organized in a tree way) that I could browse somehow like:

for table in myList:
 for field in table:
 print (field("type"))
 print (field("fk")) 

Is the Python's built-in JSON API extent wide enough to reach this expectation?

Answer 1

You seem to need help iterating over the returned object, as well as decoding the JSON.

import json
#jstr = "... that thing above ..."
# This line only decodes the JSON into a structure in memory:
obj = json.loads(jstr)
# obj, in this case, is a dictionary, a built-in Python type.
# These lines just iterate over that structure.
for ka, va in obj.iteritems():
 print ka
 for kb, vb in va.iteritems():
 print ' ' + kb
 for key, string in vb.iteritems():
 print ' ' + repr((key, string))

• Jeremy: @Blairg23 what would you do instead to process through JSON of a known format?
• Blairg23: @Jeremy If you know the data structure, just reference the keys to get the values. In this example, he's using a triple nested for loop when he could just say something like : slate_id_type = obj['slate']['id']['type']. Since we assume he's using those values for something, might as well declare them right there. Much better to be explicit anyways.
• Blairg23: @Jeremy Absolutely! Btw, don't forget to put try, except blocks around to catch any possible key value not being present. Check out docs.python.org/3.6/library/exceptions.html (change to 2.7 if you're using that) for more options for exceptions.

Answer 2

Try

obj = json.loads(jstr)

instead of

obj = json.JSONDecoder(jstr)

• Blairg23: This is the proper way to do it. json.loads() returns a Python dictionary from the JSON object.

Answer 3

The deal I guess is that you create a decoder, but never tell it to decode().

Use:

o = json.JSONDecoder().decode(jstr)

• CoolStraw: Good point. But: for o in obj: for t in o: print(t) s l a t e b a n n e d
• Thanatos: For your JSON, Python is going to return a dict (dictionary) object. By default, iterating over it gets the keys, which are strings. Iterating over a string gets you characters. You can use .iteritems() to get tuples of (key, value), or .itervalues() to get just values in your for loop on the dictionary.
• Winston Ewert: @CoolStraw, the result of decode() is a python dictionary. You'll probably want to iterator over o.items(), as it is you are iterating over the keys and then over the letters in the keys

Answer 4

The signature of JSONDecoder is

class json.JSONDecoder([encoding[, object_hook[, parse_float[, parse_int[, 
 parse_constant[, strict[, object_pairs_hook]]]]]]])

and does not accept the JSON string in the constructur. Look at its decode() method.

http://docs.python.org/library/json.html#json.JSONDecoder

Answer 5

This worked well for me, and the printing is simpler than explicitly looping through the object like in Thanatos' answer:

import json
from pprint import pprint
jstr = #### my json code above #### 
obj = json.loads(jstr)
pprint(obj)

This uses the "Data Pretty Printer" (pprint) module, the documentation for which can be found here.

Answer 6

The String you provide in the example is not valid JSON.

The last comma between two closing curly braces is illegal.

Anyway you should follow Sven's suggestion and use loads instead.

• CoolStraw: I fixed it. But I still can't browse as I'm expecting. Am gonna update my question