JSON Schema regarding use of $ref

2019-02-03 02:24发布

I understand that $ref takes a URI to a json schema to use but where does $ref : "#" point to? Does it just mean use the current schema for this block level? Or does it mean to use the root level schema defined in the root level id? Thanks

EDIT: So if I have:

"items": {
        "anyOf": [
            { "$ref": "#" },
            { "$ref": "#/definitions/schemaArray" }
        ],
        "default": {}
    }

Because it lacks an id field it will attempt to validate the instance items with the root schema first and then if that fails try to validate it with the schemaArray schema defined in the definitions schema, right?

So if I change it to:

 "items": {
            "id" : "#/items",
            "anyOf": [
                { "$ref": "#" },
                { "$ref": "#/definitions/schemaArray" }
            ],
            "default": {}
        }

Then the first subschema in anyOf array will point to the items schema itself?

EDIT #2: Okay so if I had:

 "items": {
        "id" : "itemSchema",
        "anyOf": [
            { "$ref": "#" },
            { "$ref": "#/definitions/schemaArray" }
        ],
        "default": {}
    }

and

"stringArray": {
        "type": "array",
        "items": { "$ref" : "itemSchema" },
        "minItems": 1,
        "uniqueItems": true
    }

"stringArray"'s "items" field would be validated against the above "itemsSchema"?

Also does the second $ref in 'anyOf' work by going to the root and then traversing down the path till it hits that schema? Thanks!

1条回答
三岁会撩人
2楼-- · 2019-02-03 03:18

OK: each $ref is resolved into a full URI. Once that is done, all your questions are answered by asking the question: What schema would I end up with, if I simply fetched that URI? Where the $ref is, how it was loaded, all of that is irrelevant - it's entirely dependent on the resolved URI.

The library might take some shortcuts (like caching documents so they are only fetched once, or trusting one schema to "speak for" another), but those are all implementation details.

Response to original question:

# is not special: all values of $ref are resolved as URIs relative to the current document (or the closest value of "id", if there is one).

Therefore, if you haven't used "id", then # will point to the root of the schema document. If you fetched your schema from http://example.com/schema, then a {"$ref": "#"} anywhere inside that will resolve to http://example.com/schema#, which is the document itself.

It is different when you use "id", because it changes the "base" schema against which the $ref is resolved:

{
    "type": "array",
    "items": {
        "id": "http://example.com/item-schema",
        "type": "object",
        "additionalProperties": {"$ref": "#"}
    }
}

In that example, the $ref resolves to http://example.com/item-schema#. Now, if your JSON Schema setup trusts the schema it already has, then it can re-use the value from "items".

However, the point is there is nothing special about # - it just resolves to a URI like any other.

Response to EDIT 1:

Your first example is correct.

However, your second is unfortunately not. This is because of the way that fragments resolution works for URIs: one fragment completely replaces another. When you resolve the # against the "id" value of #/items, you don't end up with #/items again - you end up with #. So in your second example, the first entry in "anyOf" will still resolve to the root of the document, just as in the first example.

Response to EDIT 2:

Assuming the document is loaded from http://example.com/my-schema, the full URIs of your two $refs are:

  • http://example.com/itemSchema#
  • http://example.com/itemSchema#/definitions/schemaArray

For the first one, the library may use the schema it already has, but it might not - after all, looking at the URIs, http://example.com/my-schema might not be trusted to accurately represent http://example.com/itemSchema.

For the second one - that's not going to work, because the "itemSchema" doesn't have a "definitions" section, so that $ref won't resolve properly at all.

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