JSON模式-如果对象*不*包含特定属性，则有效 [英] JSON schema - valid if object does *not* contain a particular property

问题描述

是否可以设置一个仍允许additionalProperties允许但 not 不匹配的JSON模式?换句话说，我需要知道是否有可能与required声明完全相反.

Is it possible to set up a JSON schema that still allows for additionalProperties but does not match if a very particular property name is present? In other words, I need to know if it's possible to have the exact opposite of the required declaration.

模式:

{
    "type": "object",
    "properties": {
        "x": { "type": "integer" }
    },
    "required": [ "x" ],
    "ban": [ "z" ] // possible?
}

匹配:

{ "x": 123 }

匹配:

{ "x": 123, "y": 456 }

不匹配:

{ "x": 123, "y": 456, "z": 789 }

推荐答案

您可以使用not关键字来完成您想做的事情.如果not模式验证，则父模式将不验证.

What you want to do can be achieved using the not keyword. If the not schema validates, the parent schema will not validate.

{
    "type": "object",
    "properties": {
        "x": { "type": "integer" }
    },
    "required": [ "x" ],
    "not": { "required": [ "z" ] }
}

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