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一尘不染

如何在Java中对JSON对象进行排序？

json

我已经寻找了一段时间，想要一种对JSON对象进行排序的方法，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",
   }
]}

并按字母顺序按值“ COMMERCIALNAME_E”得到：

{"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"
   }
]}

我找不到任何可以做到这一点的代码。谁能给我些帮助？


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2020-07-27

共1个答案

一尘不染

将这些JSON解析为“对象集合”，然后使用比较器通过您喜欢的字段对其进行排序。

例：

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);

如果要从Object创建JSON。

Person p = new Person();
p.setName("Jigar");
p.setAge(22);
String jsonStr = new com.google.gson.Gson().toJson(obj);
2020-07-27