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)

Answer 2

I also needed a converter tool and came across this. So far it seems to work pretty well. It does both JSON and YAML formats.

https://swagger-toolbox.firebaseapp.com/

Given this JSON (their sample):

{
  "id": 1,
  "name": "A green door",
  "price": 12,
  "testBool": false,
  "tags": [
    "home",
    "green"
  ]
}

it generated this:

{
    "required": [
        "id",
        "name",
        "price",
        "testBool",
        "tags"
    ],
    "properties": {
        "id": {
            "type": "number"
        },
        "name": {
            "type": "string"
        },
        "price": {
            "type": "number"
        },
        "testBool": {
            "type": "boolean"
        },
        "tags": {
            "type": "array",
            "items": {
                "type": "string"
            }
        }
    }
}

Answer 3

I know there are some tools to convert JSON to JSON schemas but, if I’m not mistaken, Swagger only has $refs to other objects definitions thus only has one level

You are mistaken. Swagger will respect any valid v4 JSON schema, as long as it only uses the supported subset.

The Schema Object...is based on the JSON Schema Specification Draft 4 and uses a predefined subset of it. On top of this subset, there are extensions provided by this specification to allow for more complete documentation.

It goes on to list the parts of JSON schema which are supported, and the bits which are not, and the bits which are extended by swagger.