使用jackson删除JSON元素

Question

我有一个特定的JSON节点，对应于import org.codehaus.jackson。JsonNode，而不是导入org.codehaus.jackson.map.JsonNode。

[
    {
        "givenName": "Jim",
        "formattedName": "jimJackson",
        "familyName": null,
        "middleName": "none",
        "honorificPrefix": "mr",
        "honorificSuffix": "none"
    },
    {
        "givenName": "john",
        "formattedName": "johnLasher",
        "familyName": null,
        "middleName": "none",
        "honorificPrefix": "mr",
        "honorificSuffix": "none"
    },
    {
        "givenName": "carlos",
        "formattedName": "carlosAddner",
        "familyName": null,
        "middleName": "none",
        "honorifiPrefix": "mr",
        "honorificSuffix": "none"
    },
    {
        "givenName": "lisa",
        "formattedName": "lisaRay",
        "familyName": null,
        "middleName": "none",
        "honorificPrefix": "mrs",
        "honorificSuffix": "none"
    },
    {
        "givenName": "bradshaw",
        "formattedName": "bradshawLion",
        "familyName": null,
        "middleName": "none",
        "honorificPrefix": "mr",
        "honorificSuffix": "none"
    },
    {
        "givenName": "phill",
        "formattedName": "phillKane",
        "familyName": null,
        "middleName": "none",
        "honorificPrefix": "mr",
        "honorificSuffix": "none"
    },
    {
        "givenName": "Gabriel",
        "formattedName": "gabrielMoosa",
        "familyName": null,
        "middleName": "none",
        "honorificPrefix": "mr",
        "honorificSuffix": "none"
    }
]

我想从上述数组的所有JSON节点中删除“familyName”和“middleName”。有没有办法做到这一点？

Answer 1

我最近想到了这个问题，因为我有一个不寻常的JSON片段，我需要删除其中的一个元素:

{
   "allan": { "score": 88 },
   "bill": { "score": 75 },
   "cassie": { "score": 96 },
   "warnings": [ { "name": "class_low_numbers", "message": "this class is has less than 10 students" }]
}

前三个元素表示一个人和一个相应的分数对象。最后一个“警告”，与分数对象不匹配，这就是我想删除的对象。

从上面gsteff的回答中，将rootNode作为起始JsonNode，我发现消除它的方法是迭代每个节点，将节点的对象映射版本添加到HashMap中，这是我想要的返回对象，除非它是“warnings”元素：

HashMap<String, ScoreDetails> results = new HashMap<String, ScoreDetails>();

ObjectMapper mapper = new ObjectMapper();
mapper.configure(MapperFeature.ACCEPT_CASE_INSENSITIVE_PROPERTIES, true);
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

Iterator<Map.Entry<String, JsonNode>> fields = rootNode.fields();
while (fields.hasNext()) {
  Map.Entry<String, JsonNode> next = fields.next();
  if (!next.getKey().equals("warnings")) {
    results.put(
        next.getKey(), mapper.treeToValue(next.getValue(), ScoreDetails.class));
  }
}

return results;

Answer 2

Jackson的ObjectMapper仅用几个步骤就给出了解决方案。

将json数据保存在一个文件中，比如“data.json”。将下面的代码复制到不带import语句的函数中，然后调用该函数。生成的JSON将写入一个新文件“data1.JSON”。

import java.io.File;
import java.io.IOException;

import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.node.ObjectNode;
ObjectMapper objectMapper = new ObjectMapper();
        JsonNode jsonNode = objectMapper.readTree(new File("data.json"));
        for (JsonNode node : jsonNode) {
            ((ObjectNode)node).remove("familyName");
            ((ObjectNode)node).remove("middleName");
        }
        objectMapper.writeValue(new File("data1.json"), jsonNode);

Answer 3

我还没有测试过这个，但是我认为类似这样的东西可以满足你的要求:

import org.codehaus.jackson.node.ObjectNode;
// ...
for (JsonNode personNode : rootNode) {
    if (personNode instanceof ObjectNode) {
        ObjectNode object = (ObjectNode) personNode;
        object.remove("familyName");
        object.remove("middleName");
    }
}

你也可以使用Jackon的原始解析API更有效地做到这一点，但代码会更混乱。