使用Gson将LinkedTreeMap解码为属性？

w80xi6nr 于 2022-12-29 发布在其他

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我有一个JSON结构，如下所示：

{
  "identifier": 1045608,
  "scientificName": "Apis mellifera Linnaeus 1758",
  "exemplar": false,
  "richness_score": 400.0,
  "dataObjects": [
    {
      "identifier": "d72801627bf4adf1a38d9c5f10cc767f",
      "dataObjectVersionID": 30073527,
      "dataType": "http://purl.org/dc/dcmitype/StillImage",
      "dataSubtype": "",
      "vettedStatus": "Trusted",
      "dataRatings": {
        "1": 0,
        "2": 0,
        "3": 4,
        "4": 0,
        "5": 6
      },
      "dataRating": 4.2,
      "mimeType": "image/jpeg",
      "created": "2009-07-12T15:13:19Z",
      "title": "Honey Bee on Mountain Mint",
      "language": "en",
      "license": "http://creativecommons.org/licenses/by/2.0/",
      "rightsHolder": "John Baker",
      "source": "https://www.flickr.com/photos/38875278@N08/3730360050/",
      "mediaURL": "https://farm3.staticflickr.com/2619/3730360050_c771a4c2cf_o.jpg",
      "agents": [
        {
          "full_name": "John Baker",
          "homepage": "http://www.flickr.com/photos/38875278@N08",
          "role": "photographer"
        },
        {
          "full_name": "Flickr: EOL Images",
          "homepage": "http://www.flickr.com/groups/encyclopedia_of_life",
          "role": "provider"
        }
      ],
    }
  ]
}

我已经定义了一个顶级类来反序列化为：

class EOLDataObjectsResponse {
    private int identifier;
    private String scientificName;
    private Boolean exemplar;
    @SerializedName("richness_score") private float richnessScore;
    private List<EOLDataObjectsTaxonConcept> taxonConcepts;
    private List<LinkedTreeMap<String, String>> dataObjects;
}

在添加dataObjects属性之前，Gson可以正确解析所有内容。

Exception in thread "main" com.google.gson.JsonSyntaxException: java.lang.IllegalStateException: Expected a string but was BEGIN_OBJECT at line 1 column 2242 path $.dataObjects[0].
    at com.google.gson.internal.bind.ReflectiveTypeAdapterFactory$Adapter.read(ReflectiveTypeAdapterFactory.java:224)
    at com.google.gson.Gson.fromJson(Gson.java:887)
    at com.google.gson.Gson.fromJson(Gson.java:825)
    at emt.eol.EOLDataObjects.query(EOLDataObjects.java:102)
    at emt.eol.EOLDataObjects.query(EOLDataObjects.java:51)
    at emt.eol.EOLDataObjects.main(EOLDataObjects.java:136)

我想做的是将这个属性解析成一个字符串-字符串对的嵌套Map列表，因为我不能保证返回到列表中的是什么，但如果有人需要访问它，我希望Gson LinkedTreeMap类能做到这一点，但显然不是我使用它的方式。
有谁能提出可能导致问题的原因或可能更好的方法吗？谢谢！

Gson

来源：https://stackoverflow.com/questions/45220294/using-gson-to-decode-a-linkedtreemap-into-a-property

1条答案

按热度按时间

lg40wkob1#

dataObjects不像其他字符串那样是一个key，它是一个json对象，你需要单独将它解析为JSON对象。

public static Map parse(final String json,final Map map){
    try {

        //Instance of JsonFactory for Object Mapper istance
        final JsonFactory factory = new JsonFactory();

        final ObjectMapper mapper = new ObjectMapper(factory);

        //create JsonNode from json String
        final JsonNode rootNode = mapper.readTree(json);

        // iterate till it fetch all parameter and value from json string 
        final Iterator<Map.Entry<String, JsonNode>> fieldsIterator = rootNode.fields();
        while (fieldsIterator.hasNext()) {

            final Map.Entry<String, JsonNode> field = fieldsIterator.next();

            //if normal json, put value to map
            map.put(field.getKey(), String.valueOf(field.getValue()));

            //if json oject again recurse parse method.
            if ((String.valueOf(field.getValue()).startsWith("{") && String.valueOf(field.getValue()).endsWith("}"))) {

                parse(String.valueOf(field.getValue()),map);

            }
            //if json array it invoke parseJsonArray 
            if (String.valueOf(field.getValue()).startsWith("[{") && String.valueOf(field.getValue()).endsWith("}]")) {

                parseJsonArray(String.valueOf(field.getValue()),map);
            }
        }
    } catch (Exception e) {
        e.printStackTrace();
    }
    return map;
   }

   public static void parseJsonArray(String jsonArray,Map<String, String> map) {
    try {
        JSONArray jsonArray1 = new JSONArray(jsonArray);
        for (int i = 0; i < jsonArray1.length(); i++) {
            JSONObject json = jsonArray1.getJSONObject(i);
            Iterator<String> keys = json.keys();

            while (keys.hasNext()) {
                String key = keys.next();
                map.put(key, String.valueOf(json.get(key)));
                System.out.println("Key :" + key + "  Value :" + String.valueOf(json.get(key)));
            }

        }
    } catch (Exception e) {
        e.printStackTrace();
    }
    }

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使用Gson将LinkedTreeMap解码为属性？

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