Convert JSON to JSON Schema draft 4 compatible with Swagger 2.0

Question

I\'ve been given some JSON files generated by a REST API with plenty of properties.

I\'ve created a Swagger 2.0 definition for this API and need to give it the corre

Answer 1

You can directly goto https://bikcrum.github.io/Swagger-JSON-Schema-In-YAML_webversion/ for online conversion.

I wrote following python script to generate JSON schema in YAML format (preserving key order) that is used in Swagger.

import json

# input file containing json file
with open('data.json') as f:
    json_data = json.load(f)

# json schema in yaml format
out = open('out.yaml','w')

def gettype(type):
    for i in ['string','boolean','integer']:
        if type in i:
            return i
    return type

def write(string):
    print(string)
    out.write(string+'\n')
    out.flush()

def parser(json_data,indent):
    if type(json_data) is dict:
        write(indent + 'type: object')
        if len(json_data) > 0:
            write(indent + 'properties:')
        for key in json_data:
            write(indent + '  %s:' % key)
            parser(json_data[key], indent+'    ')
    elif type(json_data) is list:
        write(indent + 'type: array')
        write(indent + 'items:')
        if len(json_data) != 0:
            parser(json_data[0], indent+'  ')
        else:
            write(indent + '  type: object')
    else:
        write(indent + 'type: %s' % gettype(type(json_data).__name__))

parser(json_data,'')

Update: If you want YAML with sorted keys (which is by default) use YAML library

import json
import yaml

# input file containing json file
with open('data.json') as f:
    json_data = json.load(f)

# json schema in yaml format

def gettype(type):
    for i in ['string','boolean','integer']:
        if type in i:
            return i
    return type   

def parser(json_data):
    d = {}
    if type(json_data) is dict:
        d['type'] = 'object'
        for key in json_data:
            d[key] = parser(json_data[key])
        return d
    elif type(json_data) is list:
        d['type'] = 'array'
        if len(json_data) != 0:
            d['items'] = parser(json_data[0])
        else:
            d['items'] = 'object'
        return d
    else:
        d['type'] = gettype(type(json_data).__name__)
        return d

p = parser(json_data)
with open('out.yaml','w') as outfile:
    yaml.dump(p,outfile, default_flow_style=False)