JSON 模式中的递归项

Question

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，由于

python-fastjsonschema

中的递归，我要么收到错误（使用在线验证器），要么陷入循环（机智

items.properties.children

）。

从https://json-schema.org/understanding-json-schema/structuring#recursion中的示例来看，我不清楚什么是允许递归的，什么是不允许的。不允许在

$ref

内递归

$def

，但到根锚点的

$ref

已在有效示例中列出。它们都创建了一个循环（这确实是递归的目的！）。我认为

$ref

对于非根锚点就可以了，但显然不是。

我还尝试使用未命名的锚点，如

"children": {"$ref": "#/items/properties/children"}

中所示，并且架构可以编译，但在验证时抛出错误。

我该如何建模这个模式？

Answer 1

您需要使用

#

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"]
    }
}

这是一个路过的实例。

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

您可能需要重新考虑您的

anyOf

所需的约束，因为目前对于

anyOf

，它们并不像我认为您希望的那样真正需要。试试

oneOf