4

我一直在尝试使用JSON 指针来引用和重用JSON 模式。

按照示例，我能够引用在另一个 JSON 模式中声明的特定属性，并且一切都按预期进行，但是我还没有找到一种方法来扩展基本 JSON 模式与另一个基本模式的定义，而无需显式引用每一个属性。

似乎这将是有用的，但我还没有发现迹象表明它可能与否。

想象一下基本模式things：

{
    "$schema": "http://json-schema.org/draft-04/schema#",
    "id": "http://example.com/thing.json",
    "type": "object",
    "additionalProperties": false,
    "properties": {
        "url": {
            "id": "url",
            "type": "string",
            "format": "uri"
        },
        "name": {
            "id": "name",
            "type": "string"
        }
    },
    "required": ["name"]
}

如果我想要一个更具体的person模式来重用thing我可以做的两个属性：

{
    "$schema": "http://json-schema.org/draft-04/schema#",
    "id": "http://example.com/thing/person.json",
    "type": "object",
    "additionalProperties": false,
    "properties": {
        "url": {
            "$ref": "http://example.com/thing.json#/properties/url",
        },
        "name": {
            "$ref": "http://example.com/thing.json#/properties/name",
        },
        "gender": {
            "id": "gender",
            "type": "string",
            "enum": ["F", "M"]
        },
        "nationality": {
            "id": "nationality",
            "type": "string"
        },
        "birthDate": {
            "id": "birthDate",
            "type": "string",
            "format": "date-time"
        }
    },
    "required": ["gender"]
}

但是，我发现这种方法存在两个问题：

  1. 一旦更新了超定义，依赖的模式也必须更新
  2. 手动维护所有这些引用变得繁琐/冗长
  3. 规则（如required: name）不是引用定义的一部分

有没有办法通过使用单个全局引用来获得以下有效的 JSON 模式？

{
    "$schema": "http://json-schema.org/draft-04/schema#",
    "id": "http://example.com/thing/person.json",
    "type": "object",
    "additionalProperties": false,
    "properties": {
        "url": {
            "id": "url",
            "type": "string",
            "format": "uri"
        },
        "name": {
            "id": "name",
            "type": "string"
        }
        "gender": {
            "id": "gender",
            "type": "string",
            "enum": ["F", "M"]
        },
        "nationality": {
            "id": "nationality",
            "type": "string"
        },
        "birthDate": {
            "id": "birthDate",
            "type": "string",
            "format": "date-time"
        }
    },
    "required": ["name", "gender"]
}

我尝试$ref在架构的根目录中包含，如下所示：

{
    "$schema": "http://json-schema.org/draft-04/schema#",
    "id": "http://jsonschema.net/thing/person",
    "type": "object",
    "additionalProperties": false,
    "$ref": "http://example.com/thing.json",
    "properties": {
        "gender": {/* ... */},
        "nationality": {/* ... */},
        "birthDate": {/* ... */}
    },
    "required": ["gender"]
}

这具有继承thing属性但忽略所有其他属性的效果：

gender: Additional property gender is not allowed
nationality: Additional property nationality is not allowed
birthDate: Additional property birthDate is not allowed
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1 回答 1

5

您正在寻找allOf关键字。JSON Schema 不像我们许多人习惯的那样进行继承。相反，您可以告诉它数据需要对父模式（事物）和子模式（人）都有效。

{
    "$schema": "http://json-schema.org/draft-04/schema#",
    "id": "http://example.com/thing.json",
    "type": "object",
    "properties": {
        "url": {
            "id": "url",
            "type": "string",
            "format": "uri"
        },
        "name": {
            "id": "name",
            "type": "string"
        }
    },
    "required": ["name"]
}

{
    "$schema": "http://json-schema.org/draft-04/schema#",
    "id": "http://example.com/thing/person.json",
    "allOf": [
        { "$ref": "http://example.com/thing.json" },
        {
            "type": "object",
            "properties": {
                "gender": {
                    "id": "gender",
                    "type": "string",
                   "enum": ["F", "M"]
                },
                "nationality": {
                    "id": "nationality",
                    "type": "string"
                },
                "birthDate": {
                    "id": "birthDate",
                    "type": "string",
                    "format": "date-time"
                }
            },
            "required": ["gender"]
        }
    ],
}

或者，如我所愿，更简洁地写成

{
    "$schema": "http://json-schema.org/draft-04/schema#",
    "id": "http://example.com/thing/person.json",
    "allOf": [{ "$ref": "http://example.com/thing.json" }],
    "properties": {
        "gender": {
            "id": "gender",
            "type": "string",
            "enum": ["F", "M"]
        },
        "nationality": {
            "id": "nationality",
            "type": "string"
        },
        "birthDate": {
            "id": "birthDate",
            "type": "string",
            "format": "date-time"
        }
    },
    "required": ["gender"]
}

请注意，使用这种方法，您不能使用"additionalProperties": false. 正是出于这个原因，我总是建议人们最好的做法是忽略其他属性，而不是明确禁止它们。

于 2016-01-14T01:53:21.697 回答