Question

我刚刚完成了为期6个月的编码速成课程的新手编码器。我正在研究一个java webapp来展示我的技能，而我所涉及的项目构思涉及从API中检索JSON数据，这是我们在课堂上没有学到的。我创建了POJO来匹配JSON，我正在尝试将JSON解析为java对象以存储在数据库中，但是当我运行应用程序时，我的数据库表永远不会填充数据。我怀疑问题是我的方法转换JSON，但任何反馈都非常感谢。这是我认为相关的所有代码，对不起，如果它的TMI。如果我的代码很难看，我也很抱歉，我是初学者......谢谢！

API返回JSON，如下所示：

{
    "result":{
        "status":1,
        "num_results":1,
        "total_results":500,
        "results_remaining":499,
        "matches":[{
            "match_id":3188095188,
            "match_seq_num":2784956606,
            "start_time":1495079320,
            "lobby_type":7,
            "radiant_team_id":0,
            "dire_team_id":0,
            "players":[{
                "account_id":86920222,
                "player_slot":0,
                "hero_id":18
             },{
                "account_id":61122568,
                "player_slot":1,
                "hero_id":85
             },{
                "account_id":10208661,
                "player_slot":2,
                "hero_id":13
             },{
                "account_id":106083675,
                "player_slot":132,
                "hero_id":50
             }]
         }]
    }
}

我的POJO：

@Entity
@JsonIgnoreProperties(ignoreUnknown = true)
public class Result {

    @JsonIgnore
    @Id
    @GeneratedValue
    private int id;

    @JsonProperty("status")
    private int status;

    @JsonProperty("num_results")
    private int num_results;

    @JsonProperty("total_results")
    private int total_results;

    @JsonProperty("results_remaining")
    private int results_remaining;

    @OneToMany
    @JoinColumn(name = "result_id")
    @ElementCollection(targetClass=Matches.class)
    @JsonProperty("matches")
    private List<Matches> matches;

    // getters and setters 
}

@Entity
@JsonIgnoreProperties(ignoreUnknown = true)
public class Matches {

    @Id
    @JsonProperty("match_id")
    private int match_id;

    @JsonIgnore
    @ManyToOne
    private Result result;

    @JsonProperty("match_seq_num")
    private int match_seq_num;

    @JsonProperty("start_time")
    private int start_time;

    @JsonProperty("lobby_type")
    private int lobby_type;

    @JsonProperty("radiant_team_id")
    private int radiant_team_id;

    @JsonProperty("dire_team_id")
    private int dire_team_id;

    @OneToMany
    @JoinColumn(name = "Matches_id")
    @ElementCollection(targetClass=Players.class)
    @JsonProperty("players")
    private List<Players> players;

    // getters and setters
}

@Entity
@JsonIgnoreProperties(ignoreUnknown = true)
public class Players {

    @JsonIgnore
    @Id
    @GeneratedValue
    private int id;

    @JsonIgnore
    @ManyToOne
    private Matches matches;

    @JsonProperty("account_id")
    private int account_id;

    @JsonProperty("player_slot")
    private int player_slot;

    @JsonProperty("hero_id")
    private int hero_id;

    // getters and setters
}

读取JSON并将其转换为对象的服务方法（url被审查，不希望我的API密钥公开）

public class SteamService {
    public static Result getMatchHistory(String steamid){
        Result result = new Result();
        String MatchHistoryUrl = "https:**URL**="+steamid;
        RestTemplate restTemplate = new RestTemplate();
        Result jsonresult = restTemplate.getForObject(MatchHistoryUrl, Result.class);
        return jsonresult;
    }
}

控制器

@Controller
@RequestMapping("")
public class HomeController {
    @Autowired
    private ResultsDao resultsDao;

    @RequestMapping(value = "", method = RequestMethod.GET)
    public String index(Model model){
        model.addAttribute("title", "Welcome");
        return "home/home";
    }

    @RequestMapping(value = "", method = RequestMethod.POST)
    public String processSteamIdField(@RequestParam("steamid")String steamid, Model model) {
        Result newresult = getMatchHistory(steamid);
        resultsDao.save(newresult);
        return "redirect:results";
    }
}

DAO

@Repository
@Transactional
public interface ResultsDao extends CrudRepository<Result, Integer>{
}

Answer 1

也许我的方法有点天真，但是......如果你想将JSON作为字符串存储在数据库中，那么我会使用一个对象映射器：

new ObjectMapper().writeValueAsString(myObject);

并且为了阅读JSON并将其解析为我将要执行的类：

new ObjectMapper().readValue(JSON_STRING_HERE, "utf-8"), MyPOJO.class);

此外，如果您已经使用Spring，那么您的控制器可能看起来像这样（例如，对于POST）

@RequestMapping(value = "/", method = RequestMethod.POST)
public MyPojo myController(@RequestBody MyPojo myBody) {
    myRepository.save(myBody);
}

因此，解析客户端发送到您的应用程序和控制器的JSON已经由Spring处理

无法从URL读取JSON到java对象/存储在mysql db中

1 个答案: