java - 根据 Spring Controller 方法配置 JSON 输出

Question

目前我正在使用 Jackson JSON Processor 从对象返回 JSON，但如果有更好的方法来做到这一点，我愿意接受建议。目标是根据方法/url 定义 JSON 响应中将包含哪些属性。

示例:

型号

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

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

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

Controller

@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 Jackson 注释，但它没有让我进一步了解或提出新问题。

Answer 1

您可以使用 JSON 自定义注释来忽略字段、忽略 null 、忽略 java 类 中的 empty String 到 JSON 转换，

@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。