根据JSON模式验证JSON（Java）

Question

    "$schema": "http://json-schema.org/draft-04/schema#",
    "title": "Employee",
    "description": "employee description",
    "type": "object",
    "properties": {
        "eid": {
            "description": "The unique identifier for a emp",
            "type": "integer"
        },
        "ename": {
            "description": "Name of the emp",
            "type": "string"
        },
        "qual":{
            "$ref": "#/definitions/qualification"   
        }
    },
    "definitions": {
        "qualification": 
        {
            "description": "Qualification",
            "type": "string"
        }
    }
}

这是要根据模式验证的JSON。

{
    "eid":1000,
    "ename": "mrun",
    "qualification": "BE"
}

问题是，如果我们传递了错误的数据，它将正确地验证eid和ename的类型（即整数或字符串）。对于例如：

{
    "eid":"Mrun", //should be Integer
    "ename": 72831287, //should be String
    "qualification": 98372489 //should be String
}

如果我们为限定传递了错误的类型，那么它将验证为true（即，它不验证限定的类型，可能是因为它是嵌套的）。

Answer 1

你的例子

{
    "eid":"Mrun",
    "ename": 72831287,
    "qualification": 98372489
}

与您的架构不匹配。您的架构需要像这样的对象

{
    "eid": "Mrun",
    "ename": 72831287,
    "qual": {
        "qualification": 98372489
    }
}

但是如果您只是想重用“限定”定义，那么您的模式应该类似于

"properties": {
    "eid": {
        "description": "The unique identifier for a emp",
        "type": "integer"
    },
    "ename": {
        "description": "Name of the emp",
        "type": "string"
    },
    "qualification":{
        "$ref": "#/definitions/qualification"   
    }
}