Question

目前我正在使用Jackson JSON处理器从对象返回JSON，但如果有更好的方法可以做到这一点，我愿意接受建议。目标是定义哪些属性将包含在JSON响应中，具体取决于方法/ url。

示例：

模型

class League {
  String name;
  List<Team> teams;
  ...
}

class Team {
  String name;
  String nation;
  int points;
  List<Player> players;
  ...
}

class Player {
  String name;
  Team team;
  ...
}

控制器

@Controller
public class LeagueController {

  @RequestMapping(value="/league",method = RequestMethod.GET)
  public @ResponseBody
  League getLeague() {
    //return the teams shouldn't include the player list 
  }

  @RequestMapping(value="/team/{id}",method = RequestMethod.GET)
  public @ResponseBody
  Team getTeamById(Long id) {
    //return team including the players but if possibe use the teamname inside the JSON
    //instead of the entire back reference (which producess an infinite loop.
  }
}

我查看了杰克逊的注释，但它没有让我更进一步或提出新的问题。

Answer 1

您可以使用JSON自定义注释来忽略字段，忽略null，忽略empty String到java class转换中的JSON，< / p>

@JsonIgnore
private KeyValueCollection userData;

private String participants = "";

@JsonSerialize(include = JsonSerialize.Inclusion.NON_NULL)
private boolean garbageEligible;


@JsonSerialize(include = JsonSerialize.Inclusion.NON_NULL) 
private String interactionType ;

@JsonSerialize(include = JsonSerialize.Inclusion.NON_EMPTY)
private LinkedList<InteractionInfo> interactions;

我使用了jackson 1.9。

希望这有帮助。

根据Spring Controller方法配置JSON输出

1 个答案: