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我一直在努力弄清楚如何获得一个嵌套在定义中的模式$ref。如果我对 jsonschema 对象进行硬编码，我没有任何问题，但是我需要使用嵌套模式并且我不知道里面有schema=什么validate(json{"name": "lucky","uid": -1, schema='what do I put here to reference the ref schema?')？

这是我用来隔离和测试功能的代码。

import jsonschema
from jsonschema import validate

schemas = {
    "$schema": "http://json-schema.org/draft-07/schema#",
    # "$id": "http://example.com",
    "definitions": {
        # base schema
        "uid": {
            "$id": "#uid",
            "type": "integer",
            "minimum": 1
        },
        # extended schema
        "user_record_with_ref": {
            "$id": "#user_record_with_ref",
            "type": "object",
            "properties": {
                "name": {"type": "string"},
                "uid": {
                    "allOf": [{"$ref": "#uid"}]
                }
            }
        }
    }
}

user_record_no_ref = {
    "type": "object",
    "properties": {
        "name": {"type": "string"},
        "uid": {"type": "integer", "minimum": 1}
    }
}

# I'd prefer that validation fails to prove the I am validating properly.
user_record_object = {"name": "lucky", "uid": -1}


# It is simple to validate a schema that does not use references.
def test_no_ref():
    try:
        validate(instance=user_record_object, schema=user_record_no_ref)
    except jsonschema.exceptions.ValidationError as e:
        return f'JSON Error {e}', 400

    return 'yay no problems'


def test_with_ref():
    try:
        # I want to use 'schema=user_record_with_ref' but I don't know how to access it.
        validate(instance=user_record_object, schema="")
    except jsonschema.exceptions.ValidationError as e:
        return f'JSON Error {e}', 400

    return 'yay no problems'

# Testing
print(test_no_ref())

print(test_with_ref())

我已经尝试过 schemas['definitions']['user_record_with_ref'] 但这根本不起作用。也许这需要一个 ref 解析器？我认为它不会，因为子模式位于单个模式块内。任何帮助将不胜感激。

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