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jsonschema - 从外部 JSON 模式导入所有定义

转载 作者:行者123 更新时间:2023-12-04 02:15:06 27 4
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我一直在试验 JSON Pointers引用和重用 JSON schemas .

按照示例,我能够引用在另一个 JSON 模式中声明的特定属性,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,而无需显式引用另一个基本模式的定义每一个属性(property)。

似乎这会很有用,但我还没有发现它可能与否的迹象。

想象一下基本架构 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"]
}

如果我想要更具体的 person重用 thing 的两个属性的架构我可以:
{
"$schema": "http://json-schema.org/draft-04/schema#",
"id": "http://example.com/thing/person.json",
"type": "object",
"additionalProperties": false,
"properties": {
"url": {
"$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:
  • 一旦更新了超定义,依赖模式也必须更新
  • 手动维护所有这些引用变得繁琐/冗长
  • 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"]
    }

    我试过包括 $ref在模式的根目录中,如下所示:
    {
    "$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的作用属性,但忽略所有其他属性:
    gender: Additional property gender is not allowed
    nationality: Additional property nationality is not allowed
    birthDate: Additional property birthDate is not allowed

    最佳答案

    您正在寻找 allOf关键词。 JSON Schema 不像我们很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。

    关于jsonschema - 从外部 JSON 模式导入所有定义,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34771998/

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