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java - Gson - 序列化时避免不必要的包装类

转载 作者:塔克拉玛干 更新时间:2023-11-02 19:23:09 24 4
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我有一个 Json 字符串:

{
"Locations":{
"Location":[{
"elevation":"100",
"latitude":"0",
"longitude":"0",
"name":"Ocean"
},
...
]}
}

我想使用 Gson 将其序列化为 Location 类的数组:

public class Location {
public double elevation;
public double latitude;
public double longitude;
public String name;
}

而不是创建一个 Locations 类和一个带有 Locations 字段的包装类。这是可能的吗?

最佳答案

您可以获得Locaitons 对象,然后是Location 数组。下面的代码是纯 Gson:

String jsonAsText = "{\"Locations\":{\"Location\":[{\"elevation\":\"100\",\"latitude\":\"0\",\"longitude\":\"0\",\"name\":\"Ocean\"},{\"elevation\":\"100\",\"latitude\":\"0\",\"longitude\":\"0\",\"name\":\"Ocean\"}]}}";

GsonBuilder gsonBuilder = new GsonBuilder();
Gson gson = gsonBuilder.create();

java.lang.reflect.Type listType = new com.google.gson.reflect.TypeToken<List<Location>>(){}.getType();

JsonArray locationJsonList = (JsonArray) ((JsonObject) gson.fromJson(jsonAsText, JsonObject.class).get("Locations")).get("Location");
List<Location> locatioList = gson.fromJson(locationJsonList, listType);

关于java - Gson - 序列化时避免不必要的包装类,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26084892/

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