如何在java中对JSON对象进行排序？

Question

我一直在寻找一段时间，想要一种方法来对 JSON 对象进行排序，如下所示：

 {"results": [
  {
    "layerId": 5,
    "layerName": "Pharmaceutical Entities",
    "attributes": {
      "OBJECTID": "35",
      "FACILITYTYPE": "Pharmacy",
      "FACILITYSUBTYPE": "24 Hr Pharmacy",
      "COMMERCIALNAME_E": "SADD MAARAB PHARMACY"
      },
    "geometryType": "esriGeometryPoint",
   },
  {
    "layerId": 5,
    "layerName": "Pharmaceutical Entities",
    "attributes": {
      "OBJECTID": "1",
      "FACILITYTYPE": "Pharmacy",
      "FACILITYSUBTYPE": "24 Hr Pharmacy",
      "COMMERCIALNAME_E": "GAYATHY HOSPITAL  PHARMACY"
    },
    "geometryType": "esriGeometryPoint",
  },
     {
    "layerId": 5,
    "layerName": "Pharmaceutical Entities",
    "attributes": {
      "OBJECTID": "255",
      "FACILITYTYPE": "Pharmacy",
      "FACILITYSUBTYPE": "24 Hr Pharmacy",
      "COMMERCIALNAME_E": "AL DEWAN PHARMACY"
      },
    "geometryType": "esriGeometryPoint",
   }
]}

并按“COMMERCIALNAME_E”的值按字母顺序排序以获得：

 {"results": [
   {
    "layerId": 5,
    "layerName": "Pharmaceutical Entities",
    "attributes": {
      "OBJECTID": "255",
      "FACILITYTYPE": "Pharmacy",
      "FACILITYSUBTYPE": "24 Hr Pharmacy",
      "COMMERCIALNAME_E": "AL DEWAN PHARMACY"
      },
    "geometryType": "esriGeometryPoint"
   },
  {
    "layerId": 5,
    "layerName": "Pharmaceutical Entities",
    "attributes": {
      "OBJECTID": "1",
      "FACILITYTYPE": "Pharmacy",
      "FACILITYSUBTYPE": "24 Hr Pharmacy",
      "COMMERCIALNAME_E": "GAYATHY HOSPITAL  PHARMACY"
       },
    "geometryType": "esriGeometryPoint"
   },
   {
    "layerId": 5,
    "layerName": "Pharmaceutical Entities",
    "attributes": {
      "OBJECTID": "35",
      "FACILITYTYPE": "Pharmacy",
      "FACILITYSUBTYPE": "24 Hr Pharmacy",
      "COMMERCIALNAME_E": "SADD MAARAB PHARMACY"
      },
    "geometryType": "esriGeometryPoint"
   }
]}

我找不到任何可以执行此操作的代码。谁能给我一些帮助？

原文由 Venkatesh Goud 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

将这些 JSON 解析为对象集合，并使用比较器使用您的首选字段对其进行排序。

• 使用 GSON 将其解析为对象集合

例子：

 import com.google.gson.Gson;

class Person {
  private int age;
  private String name;
}

String json = "{'age':22,'name':'Jigar'}";
Gson gson = new Gson();
TestJsonFromObject obj = gson.fromJson(json, Person.class);

如果你想从对象创建 JSON。

 Person p = new Person();
p.setName("Jigar");
p.setAge(22);
String jsonStr = new com.google.gson.Gson().toJson(obj);

原文由 jmj 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 2

我使用 JSON 简单 API 对其进行排序。这是我的代码：

 import java.io.FileReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;

import org.json.simple.JSONArray;
import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;

public class SortJSON {

public static void main(String[] args) {
    JSONParser parser = new JSONParser();
    try {
        JSONObject o = (JSONObject) parser.parse(new FileReader("test3.json"));
        JSONArray array = (JSONArray) o.get("results");
        ArrayList<JSONObject> list = new ArrayList<>();

        for (int i = 0; i < array.size(); i++) {
            list.add((JSONObject) array.get(i));
        }
        Collections.sort(list, new MyJSONComparator());

        for (JSONObject obj : list) {
            System.out.println(((JSONObject) obj.get("attributes")).get("OBJECTID"));
        }
    } catch (Exception e) {
        e.printStackTrace();
    }
}

}

class MyJSONComparator implements Comparator<JSONObject> {

@Override
public int compare(JSONObject o1, JSONObject o2) {
    String v1 = (String) ((JSONObject) o1.get("attributes")).get("COMMERCIALNAME_E");
    String v3 = (String) ((JSONObject) o2.get("attributes")).get("COMMERCIALNAME_E");
    return v1.compareTo(v3);
}

}

原文由 Sarneet Kaur 发布，翻译遵循 CC BY-SA 3.0 许可协议