我在JSON的形式的一些数据,以及使用该GSON库来解析为一个Java对象中的代码的后面部分中使用了。的JSON具有嵌套的对象,这似乎并没有被正确地得到解析,我无法找出原因,如所需外部对象正在转换。下面是我在看的JSON数据的例子:
{
"title":"Emergency Services Headquarters",
"description":"",
"cid":"C70856",
"building_id":"4714",
"building_number":"3542",
"campus_code":"20",
"campus_name":"Busch",
"location":{
"name":"Emergency Services Headquarters",
"street":"129 DAVIDSON ROAD",
"additional":"",
"city":"Piscataway",
"state":"New Jersey",
"state_abbr":"NJ",
"postal_code":"08854-8064",
"country":"United States",
"country_abbr":"US",
"latitude":"40.526306",
"longitude":"-74.461470"
},
"offices":[
"Emergency Services"
]
}
我用codebeautify创建的JSON(一切都内Building.java)所需的Java对象类:
public class Building {
private String title;
private String description;
private String cid;
private String building_id;
private String building_number;
private String campus_code;
private String campus_name;
Location LocationObject;
ArrayList < Object > offices = new ArrayList < Object > ();
//Setters and getters have been omitted
}
class Location {
private String name;
private String street;
private String additional;
private String city;
private String state;
private String state_abbr;
private String postal_code;
private String country;
private String country_abbr;
private String latitude;
private String longitude;
//Setters and getters have been omitted
}
下面是我使用解析JSON,其中变量JSON是用于该方法的输入参数的代码:
Gson obj = new Gson();
JsonArray buildingsArray = new JsonArray();
JsonParser parser = new JsonParser();
JsonElement jsonElement = parser.parse(json);
buildingsArray = jsonElement.getAsJsonArray();
for (int i = 0; i < buildingsArray.size(); i++)
Building building = obj.fromJson(buildingsArray.get(i), Building.class);
当我打电话的方法,如building.getTitle()或building.getCid(),我得到了适当的值,但是当我做building.getLocation()(其中位置是一个单独的对象),代码返回NULL。我一直没能弄明白,是它与GSON工作方式的问题?还是我在做我的代码出问题了吗?
首先,改变:
Location LocationObject;
至:
private Location location;
而且,您可以deserialise JSON
容易得多:
Gson gson = new GsonBuilder().create();
Building building = gson.fromJson(json, Building.class);
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