我想将以下JSON字符串转换为Java对象:
String jsonString = "{ "libraryname":"My Library", "mymusic":[{"Artist Name":"Aaron","Song Name":"Beautiful"}, {"Artist Name":"Britney","Song Name":"Oops I did It Again"}, {"Artist Name":"Britney","Song Name":"Stronger"}]}"
我的目标是轻松访问它,例如:
(e.g. MyJsonObject myobj = new MyJsonObject(jsonString) myobj.mymusic[0].id would give me the ID, myobj.libraryname gives me "My Library").
我听说过Jackson,但是由于涉及“ mymusic”列表,我不知道如何使用它来适应我拥有的json字符串,因为它不仅仅是键值对。如果杰克逊不是最好的选择,我该如何用杰克逊完成?
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);