如何使用杰克逊创建嵌套的 json 对象

Question

我正在使用hibernate从我的数据库中查询用户记录，如下所示：

String hql = "SELECT U FROM User U";
List<User> users = this.em.createQuery(hql).getResultList();

如何将返回的数据转换成如下格式的json：

如果没有找到用户，我希望我的 json 如下所示：

我想在创建 json 数据之前对检索到的数据进行一些操作。因此，我想避免直接传递列表，如下所示：

 String data = mapper.writerWithDefaultPrettyPrinter().writeValueAsString(users);

Answer 1

如果您使用的是 Jackson 库，您应该实现一个简单的 DTO 对象，其中包含一个 User / UserDto 列表，一个带有自定义 msg 属性的空列表的存根，然后根据可用结果创建适当的实例.

您还可以使用@JsonPropertyOrder 注解在生成的 JSON 中设置属性的顺序。

public class UserDto {
    private final String name;
    private final String email;

    public UserDto(String name, String email) {
        this.name = name;
        this.email = email;
    }
    public String getName() { return this.name; }
    public String getEmail() { return this.email; }
}

@JsonPropertyOrder({ "usersFound", "users" })
public class ListUserDto {
    private final List<UserDto> users;

    public ListUserDto(List<UserDto> users) {
        this.users = users;
    }

    public List<UserDto> getUsers() {
        return new ArrayList<>(users);
    }

    @JsonProperty("usersFound")
    public boolean isUsersFound() {
        return true;
    }
}

@JsonPropertyOrder({ "usersFound", "msg" })
@JsonIgnoreProperties(value = {"users"})
public class EmptyListUserDto extends ListUserDto {
    public EmptyListUserDto() {
        super(Collections.emptyList());
    }

    public String getMsg() {
        return "No User Found";
    }

    @JsonProperty("usersFound")
    public boolean isUsersFound() {
        return false;
    }
}


// -------------
public ListUserDto buildUserListJson(List<User> users) {
//    String hql = "SELECT U FROM User U";
//    List<User> users = this.em.createQuery(hql).getResultList();

    if (null == users || users.isEmpty()) {
        return new EmptyListUserDto();
    }
    return new ListUserDto(
            users.stream()
                 .map(u -> new UserDto(u.getName(), u.getEmail()))
                 .collect(Collectors.toList())
    );
}

测试代码：

ObjectWriter writer = new ObjectMapper().writerWithDefaultPrettyPrinter();

System.out.println(writer.writeValueAsString(buildUserListJson(
    List.of(new User("jack", "jack@mail.com"), new User("john", "john@mail.com"))
)));

System.out.println("null -> " + writer.writeValueAsString(buildUserListJson(null)));

System.out.println("empty -> " + writer.writeValueAsString(buildUserListJson(Collections.emptyList())));

输出：

{
  "usersFound" : true,
  "users" : [ {
    "name" : "jack",
    "email" : "jack@mail.com"
  }, {
    "name" : "john",
    "email" : "john@mail.com"
  } ]
}
null -> {
  "usersFound" : false,
  "msg" : "No User Found"
}
empty -> {
  "usersFound" : false,
  "msg" : "No User Found"
}

Answer 2

你可以简单的一步一步构建json：

首先我建议在User 类中创建一个方法，将用户的姓名和电子邮件作为json。

public class User{

    ...
 
    public String getJsonUserEmail(){

        return "{\"name\": \"" + this.getName() + "\", " +
                "\"email\": \"" + this.getEmail() + "\"}"

    }

    ...

}

比将所有用户转换为json并构建最后一个json。

public static void main(String[] args) {

    String finalJson = "{";

    if(!users.isEmpty()){

        finalJson += "\"usersFound\": true, ";
                
        //transform every user to json
        //{"name": "name1", "email": "email1"}
        List<String> usersJson = users.stream()
                                      .map(e -> e.getJsonUserEmail())
                                      .collect(Collectors.toList());

        //join all jsons
        //{"name": "name1", "email": "email1"}, {"name": "name2", "email": "email2"}, ...
        String usersJsonJoin = String.join(", ", usersJson);

        finalJson += "\"users\": [" + usersJsonJoin + "]";

    }else{

        finalJson += "\"usersFound\": false, ";
        finalJson += "\"msg\": \"No users found\"";

    }

    finalJson += "}";

}