jsonschema

C# library for converting json schema to sample JSON

眉间皱痕 提交于 2020-08-03 02:48:55
问题 I'm looking for a C# library that will generate a valid JSON object based on a given JSON Schema. I'd like to produce a very simple JSON sample just like how Swagger does it: I've seen some JavaScript libraries like JSON Schema Faker, but I need a C#/.Net library where I can generate sample JSON in my backend code. 回答1: Ok, it is super simplistic and doesn't take into account many factors of JSON schema, but it might be a good enough starting point for you. It also depends on the JsonSchema

C# library for converting json schema to sample JSON

与世无争的帅哥 提交于 2020-08-03 02:48:49
问题 I'm looking for a C# library that will generate a valid JSON object based on a given JSON Schema. I'd like to produce a very simple JSON sample just like how Swagger does it: I've seen some JavaScript libraries like JSON Schema Faker, but I need a C#/.Net library where I can generate sample JSON in my backend code. 回答1: Ok, it is super simplistic and doesn't take into account many factors of JSON schema, but it might be a good enough starting point for you. It also depends on the JsonSchema

creating mongoengine class dynamicly from json schema

|▌冷眼眸甩不掉的悲伤 提交于 2020-08-02 06:53:37
问题 i'm looking for a way to take my json schema and dynamically create a mongoengine class at runtime. for example: the mongoengine class written below class user(Document): _id = StringField(required=False) # mongodb id name = StringField(required=True) # user name email= StringField(required=False,regex="^[a-zA-Z0-9]*@mydomain.com$") # user email will be the same as the dynamically generated at runtime class generated from this schema { "type":"object", "properties":{ "_id" : {"type":"string"}

creating mongoengine class dynamicly from json schema

夙愿已清 提交于 2020-08-02 06:53:07
问题 i'm looking for a way to take my json schema and dynamically create a mongoengine class at runtime. for example: the mongoengine class written below class user(Document): _id = StringField(required=False) # mongodb id name = StringField(required=True) # user name email= StringField(required=False,regex="^[a-zA-Z0-9]*@mydomain.com$") # user email will be the same as the dynamically generated at runtime class generated from this schema { "type":"object", "properties":{ "_id" : {"type":"string"}

create Ref-resolver from $ref in jsonschema in python draft7

安稳与你 提交于 2020-06-17 12:56:52
问题 I have json Schema { "$id": "d:/documents/schemaFiles/WLWorkProduct/1", "$schema": "http://json-schema.org/draft-07/schema#", "title": "WLWorkProduct", "description": "", "type": "object", "properties": { "RID": { "description": "resource type.", "type": "string" }, "Data": { "type": "object", "properties": { "IndividualTypeProperties": { "allOf": [ { "$ref": "d:/documents/schemaFiles/WorkProduct/1" }, { "type": "object", "properties": { "WID": { "type": "string", "description": "WID" } } } ]

create Ref-resolver from $ref in jsonschema in python draft7

≯℡__Kan透↙ 提交于 2020-06-17 12:56:31
问题 I have json Schema { "$id": "d:/documents/schemaFiles/WLWorkProduct/1", "$schema": "http://json-schema.org/draft-07/schema#", "title": "WLWorkProduct", "description": "", "type": "object", "properties": { "RID": { "description": "resource type.", "type": "string" }, "Data": { "type": "object", "properties": { "IndividualTypeProperties": { "allOf": [ { "$ref": "d:/documents/schemaFiles/WorkProduct/1" }, { "type": "object", "properties": { "WID": { "type": "string", "description": "WID" } } } ]

django postgresql json field schema validation

情到浓时终转凉″ 提交于 2020-06-09 17:32:05
问题 I have a django model with a JSONField (django.contrib.postgres.fields.JSONField) Is there any way that I can validate model data against a json schema file? (pre-save) Something like my_field = JSONField(schema_file=my_schema_file) 回答1: I wrote a custom validator using jsonschema in order to do this (Django 1.11, Python 3.6). project/validators.py import django from django.core.validators import BaseValidator import jsonschema class JSONSchemaValidator(BaseValidator): def compare(self, a, b)

django postgresql json field schema validation

半世苍凉 提交于 2020-06-09 17:31:30
问题 I have a django model with a JSONField (django.contrib.postgres.fields.JSONField) Is there any way that I can validate model data against a json schema file? (pre-save) Something like my_field = JSONField(schema_file=my_schema_file) 回答1: I wrote a custom validator using jsonschema in order to do this (Django 1.11, Python 3.6). project/validators.py import django from django.core.validators import BaseValidator import jsonschema class JSONSchemaValidator(BaseValidator): def compare(self, a, b)