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:
required: name
)不是引用定义 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/