Question

我已经基于我的存储库和服务在数据库中保存了一个数据。我想通过仅更改玩家ID来与邮递员保存另一个数据，但它不是创建新的实体数据，而是更新现有的实体数据我的问题是如何在找到现有ID时通过服务更新数据。但是，当找到新ID时，会将新数据保存到databse中。

这是我的仓库：

@Repository
@Transactional
public interface CricketPlayerRepository extends CrudRepository<CricketPlayer,String>  {
    Optional<CricketPlayer> findCricketPlayerByName(String name);


}

这是我的服务

@Service
public class CricketPlayerService {

    private CricketPlayerRepository cricketPlayerRepository;


    public CricketPlayerService(CricketPlayerRepository cricketPlayerRepository) {
        super();
        this.cricketPlayerRepository = cricketPlayerRepository;
    }

    public CricketPlayerService() {
    }

    public Optional<CricketPlayer> getPlayerByName(String name){
        return cricketPlayerRepository.findCricketPlayerByName(name);
    }

    public CricketPlayer save(CricketPlayer cricketPlayer){
        Optional<CricketPlayer> id = cricketPlayerRepository.findById(cricketPlayer.getPlayerId());
        if (id.isPresent()){
            //code here
        }
//        if (entityManager.isNew(cricketPlayer)) {
//            em.persist(cricketPlayer);
//            return cricketPlayer;
//        } else {
//            return em.merge(cricketPlayer);
//        }
        return cricketPlayerRepository.save(cricketPlayer);
    }

    public Iterable<CricketPlayer> findAllPlayers() {
        return cricketPlayerRepository.findAll();
    }

    public Optional<CricketPlayer> findPlayersById(String id) {
        return cricketPlayerRepository.findById(id);
    }

}

Answer 1

休眠（和其他框架）中的

保存和更新操作基于id值。如果ID存在，则合并（更新）实体，否则保存新实例。因此，在这种情况下无法做到这一点。

Answer 2

1）如果PlayerId是主键ID，则您将调用merge（entity）。如果存在PlayerId，它将进行更新，否则它将创建新记录。

2）如果PlayrerId不是主键ID。最佳做法是避免将PlayerId作为主键。

在邮递员中，您应该将数据库表的主键ID与PlayerId一起传递。然后，您调用merge（entity）。它将负责基于主键的创建或更新 id是否为空。

例如，如果您在请求中传递了主键ID，则如下所示。

    Entity e = new Entity();
    if (entityFromRequest.getId() != null){ //this is the db id from request
        //get the entity from db and then change the state   
        e = entity from db by using id
    }
    e.setPlayerId = entityFromRequest.getPlayerId
    merge(e);  // it will create or update record

在jpa CRUD存储库中保存和更新

2 个答案: