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Java 不搜索子节点的JsonNode findValue_Java_Json_Jackson - Fatal编程技术网

Java 不搜索子节点的JsonNode findValue

Java 不搜索子节点的JsonNode findValue,java,json,jackson,Java,Json,Jackson,我有一个结构如下的资源: "activity": { "activity_type": "Like", "activity_id": "123456", "object_id": "", "product_id":"", "reference_activity": { "activity_type": "Rating",

我有一个结构如下的资源:

"activity": {
            "activity_type": "Like",
            "activity_id": "123456",
            "object_id": "",
            "product_id":"",
            "reference_activity": {
                    "activity_type": "Rating",
                    "activity_id": "789456",
                    "object_id": "rating/5",
                    "product_id": "product.v1.13445123",
                    "timestamp": 1433447044000
            },
            "timestamp": 1435790640115
}
问题是资源可能包含同一类的对象,即引用另一个活动。因此,它们具有相似的值名称。我试图使用Jackson的JsonNode.findValue方法解析这个JSON结构,但是findValue方法也进行嵌套搜索。因此,如果我执行findValueobject_id,它将永远不会返回空字符串。相反,它将始终返回额定值/5

有没有办法将findValue方法限制为立即数 儿童 或者,是否有一种方法可以从 父结构,但仍然保留JsonNode对象,以便可以单独解析它们?
如果不需要递归查找,只需使用path或get方法?

这就是我要找的!谢谢