Python解码JSON

Question

我有以下json:

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

我想找出最好的解码方式,让它具有易于浏览的python对象呈现方式.

我试过了:

import json

jstr = #### my json code above #### 
obj = json.JSONDecoder().decode(jstr)

for o in obj:
  for t in o: 
    print (o)

但我得到:

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

我不明白这是什么交易.理想的是一棵树(甚至以树的方式组织的列表),我可以以某种方式浏览:

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

Python的内置JSON API范围是否足以达到此预期？

Answer 1

您似乎需要帮助迭代返回的对象,以及解码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))

Answer 2

尝试

obj = json.loads(jstr)

代替

obj = json.JSONDecoder(jstr)