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user3758232
Asked: 2024-03-02 00:28:32 +0800 CST

JSON 架构中的递归项

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我正在设计一个带有递归元素的 JSON 模式。该元素不在根，而是在子模式中：

{
    "$schema": "https://json-schema.org/draft/2020-12/schema",
    "$id": "http://example.org/json-schema/struct.schema.json",
    "type": "array",
    "items": {
        "title": "Structure",
        "type": "object",
        "properties": {
            "id": {
                "title": "ID",
                "type": "string"
            },
            "label": {
                "title": "Label",
                "type": "string"
            },
            "children": {
                "$anchor": "children",
                "title": "Children",
                "type": "array",
                "items": {
                    "title": "Structure Item",
                    "type": "object",
                    "properties": {
                        "id": {
                            "title": "ID",
                            "type": "string"
                        },
                        "href": {
                            "title": "Reference",
                            "type": "string"
                        },
                        "label": {
                            "title": "Structure Item Label",
                            "type": "string"
                        },
                        "sort_label": {
                            "title": "Sort Label",
                            "type": "string"
                        },
                        "children": {"$ref": "children"}
                    }
                },
                "anyOf": [
                    {"required": ["id", "href"]},
                    {"required": ["id", "children"]}
                ]
            }
        },
        "required": ["id", "label", "children"]
    }
}

当我尝试编译模式时，python-fastjsonschema由于items.properties.children.

从https://json-schema.org/understanding-json-schema/structuring#recursion中的示例中，我不清楚什么是允许递归的，什么是不允许的。不允许$ref在 s 内递归s，但s 到根锚点会在有效示例中列出。它们都创建了一个循环（这确实是递归的目的！）。我认为对于非根锚点来说是可以的，但显然不是。$def$ref$ref

我还尝试使用未命名的锚点，如"children": {"$ref": "#/items/properties/children"}，并且模式编译，但在验证时抛出错误。

我该如何建模这个模式？

recursion

1 个回答

  1. Best Answer

    您需要用#符号引用锚点。

    {
    "$schema": "https://json-schema.org/draft/2020-12/schema",
    "$id": "http://example.org/json-schema/struct.schema.json",
    "type": "array",
    "items": {
        "title": "Structure",
        "type": "object",
        "properties": {
            "id": {
                "title": "ID",
                "type": "string"
            },
            "label": {
                "title": "Label",
                "type": "string"
            },
            "children": {
                "$anchor": "children",
                "title": "Children",
                "type": "array",
                "items": {
                    "title": "Structure Item",
                    "type": "object",
                    "properties": {
                        "id": {
                            "title": "ID",
                            "type": "string"
                        },
                        "href": {
                            "title": "Reference",
                            "type": "string"
                        },
                        "label": {
                            "title": "Structure Item Label",
                            "type": "string"
                        },
                        "sort_label": {
                            "title": "Sort Label",
                            "type": "string"
                        },
                        "children": {"$ref": "#children"}
                    }
                },
                "anyOf": [
                    {"required": ["id", "href"]},
                    {"required": ["id", "children"]}
                ]
            }
        },
        "required": ["id", "label", "children"]
    }
}

    这是一个路过的例子。

    [
    {
        "id": "",
        "label": "",
        "children": [
            {
                "id": "",
                "children": []
            },
            {
                "id": "",
                "href": ""
            }
        ]
    }
]

    您可能需要重新考虑anyOf所需的约束，因为目前anyOf，它们并不像我认为您希望的那样真正需要。尝试oneOf

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