Question

我想定义一个像这样的对象，以映射到类似java Map的东西（但不完全是，因为这也需要与typescript接口）。最终，我想弄清楚的是如何在另一个对象中声明<T>类型的东西（比如说Dict<MyObj>）（如果这是可能的话）：

（这显然不是一个有效的json架构）

{
    "type": "object",
    "javaType": "some.package.base.DictKeyValuePair",
    "properties": {
        "key": { "type": "string" },
        "value": { "type": "<T>" }
    }
}

将在其他json模式中引用，如下所示：

{
    "type": "object",
    "javaType": "some.package.base.Dict",
    "properties": {
        "key": { "type": "string" },
        "value": { "type": "<T>" },
        "keValuePairs": {
            "type": "array",
            "items": {
                "$ref": "DictKeyValuePair.json",
                "genericType": "<T>"
            }
        }
    }
}

最终用于与此类似的对象：

{
    "type": "object",
    "javaType": "some.package.SomeObject",
    "properties": {
        "someDict": {
            "$ref": "Dict.json",
            "genericType": "SomeObjectOrSimpleType"
        }
    }
}

所以...这在json架构中是否可行？

Answer 1

使用json-schema可以做类似的事情，但你需要定义模式，看看这个样本：

http://json-schema.org/learn/examples/address.schema.json

{
"$id": "https://example.com/address.schema.json",
"$schema": "http://json-schema.org/draft-07/schema#",
"description": "An address similar to http://microformats.org/wiki/h-card",
"type": "object",
"properties": {
"post-office-box": {
"type": "string"
},
"extended-address": {
"type": "string"
},
"street-address": {
"type": "string"
},
"locality": {
"type": "string"
},
"region": {
"type": "string"
},
"postal-code": {
"type": "string"
},
"country-name": {
"type": "string"
}
},
"required": [
"locality",
"region",
"country-name"
],
"dependencies": {
"post-office-box": [
"street-address"
],
"extended-address": [
"street-address"
]
}
}

Json通用对象类型

1 个答案: