Question

我正在使用 Spring 开发我的 Web 应用程序后端。特别是，我的应用程序管理有关足球队及其球员的数据。

我的应用程序与第三方 REST API 交互以获取球队和球员数据。

对于团队，我创建了一个 Team 实体，如下所示：

@Data
@Table(name = "team")
@Entity
public class Team {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id", nullable = false)
    private Long id;
    private String name;
    private String logoUrl;
    @OneToMany(fetch = FetchType.EAGER, mappedBy = "team")
    private Set<Player> players;
}

但是，来自 API 的响应具有特定结构，并且在“响应”节点中包含一组团队。

这是响应的结构：

{
   "get":"teams",
   "parameters":{
      "league":"135",
      "season":"2020"
   },
   "errors":[
      
   ],
   "results":20,
   "paging":{
      "current":1,
      "total":1
   },
   "response":[
      {
         "team":{
            "id":487,
            "name":"Lazio",
            "country":"Italy",
            "founded":1900,
            "national":false,
            "logo":"https:\/\/media.api-sports.io\/football\/teams\/487.png"
         },
         "venue":{
            "id":910,
            "name":"Stadio Olimpico",
            "address":"Viale dei Gladiatori, 2 \/ Via del Foro Italico",
            "city":"Roma",
            "capacity":68530,
            "surface":"grass",
            "image":"https:\/\/media.api-sports.io\/football\/venues\/910.png"
         }
      },
      {
         "team":{
            "id":488,
            "name":"Sassuolo",
            "country":"Italy",
            "founded":1922,
            "national":false,
            "logo":"https:\/\/media.api-sports.io\/football\/teams\/488.png"
         },
         "venue":{
            "id":935,
            "name":"MAPEI Stadium - Citt\u00e0 del Tricolore",
            "address":"Piazza Azzuri d&apos;Italia, 1",
            "city":"Reggio nell&apos;Emilia",
            "capacity":23717,
            "surface":"grass",
            "image":"https:\/\/media.api-sports.io\/football\/venues\/935.png"
         }
      },
      ... // Other team objects
   ]
}

如何使用 Jackson 库解析答案以获得 List<Team>？

Answer 1

您应该为 Jackson 创建匹配结果结构的类，然后将这些类的实例转换为您的 Team 类。对 JPA 实体和 Jackson 反序列化使用相同的类是个坏主意。

有些在线服务允许生成这样的类。例如这个 https://json2csharp.com/json-to-pojo 生成的类是这样的：

// import com.fasterxml.jackson.databind.ObjectMapper; // version 2.11.1
// import com.fasterxml.jackson.annotation.JsonProperty; // version 2.11.1
/* ObjectMapper om = new ObjectMapper();
Root root = om.readValue(myJsonString), Root.class); */
public class Parameters{
    public String league;
    public String season;
}

public class Paging{
    public int current;
    public int total;
}

public class Team{
    public int id;
    public String name;
    public String country;
    public int founded;
    public boolean national;
    public String logo;
}

public class Venue{
    public int id;
    public String name;
    public String address;
    public String city;
    public int capacity;
    public String surface;
    public String image;
}

public class Response{
    public Team team;
    public Venue venue;
}

public class Root{
    public String get;
    public Parameters parameters;
    public List<Object> errors;
    public int results;
    public Paging paging;
    public List<Response> response;
}

Answer 2

正如@Sankozi 所说，您可以对您的 java pojo 进行建模以进行 json 反序列化。然后使用 ObjectMapper 进行反序列化，例如：

ObjectMapper mapper = new ObjectMapper();                 
CollectionType javaType = mapper.getTypeFactory()
  .constructCollectionType(List.class, Response.class);                 
List<Response> asList = mapper.readValue(jsonArray, javaType);                 
List<Team> teams = asList.stream()     
  .flatMap(response -> response.getTeam())                         
  .collect(Collectors.toList());

使用 Jackson 反序列化复杂的 JSON 响应

2 个答案: