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TA贡献2039条经验 获得超7个赞
为此,无需与GSON一起使用;杰克逊可以做普通的地图/列表:
ObjectMapper mapper = new ObjectMapper();
Map<String,Object> map = mapper.readValue(json, Map.class);
或更方便的JSON树:
JsonNode rootNode = mapper.readTree(json);
顺便说一句,没有理由为什么您不能实际创建Java类并更方便地执行(IMO):
public class Library {
@JsonProperty("libraryname")
public String name;
@JsonProperty("mymusic")
public List<Song> songs;
}
public class Song {
@JsonProperty("Artist Name") public String artistName;
@JsonProperty("Song Name") public String songName;
}
Library lib = mapper.readValue(jsonString, Library.class);
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TA贡献1836条经验 获得超3个赞
Gson gson = new Gson();
JsonParser parser = new JsonParser();
JsonObject object = (JsonObject) parser.parse(response);// response will be the json String
YourPojo emp = gson.fromJson(object, YourPojo.class);
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