java - 使用 JPA 的 Java Spring Boot 中的 Column X Binary (255) 值太长

Question

我正在尝试创建一个具有多个站点的游戏，例如金字塔或寺庙。我对所有网站都遇到相同的错误，因此我仅使用 Temple 作为示例。我想做的是通过创建新站点并将其分配给游戏来初始化游戏板，反之亦然。在站点类中设置游戏工作正常，但在父“Game.java”中设置站点会引发以下错误:

2017-04-13 17:23:10.183 WARN 5764 --- [ main] o.h.engine.jdbc.spi.SqlExceptionHelper : SQL Error: 22001, SQLState: 22001 2017-04-13 17:23:10.183 ERROR 5764 --- [ main] o.h.engine.jdbc.spi.SqlExceptionHelper : Value too long for column "TEMPLE BINARY(255)": "X'aced00057372002d63682e757a682e6966692e7365616c2e736f707261667331372e656e746974792e73697465732e54656d706c65bfa968665c9a87790200... (2722)"; SQL statement: update game set burial_chamber=?, current_player=?, market=?, name=?, obelisk=?, ownerid=?, pyramid=?, shipyard=?, status=?, temple=? where game_id=? [22001-191] 2017-04-13 17:23:10.185 INFO 5764 --- [ main] o.h.e.j.b.internal.AbstractBatchImpl : HHH000010: On release of batch it still contained JDBC statements

import javax.persistence.*;
import java.io.Serializable;
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.List;

@Entity
public class Temple implements Serializable {

    @Column
    private boolean isDockEmpty = true;

    @Id
    @GeneratedValue
    @Column(name = "id", updatable = false, nullable = false)
    private long id;

    @ElementCollection
    private List<Color> stones = new ArrayList<Color>();

    public List<Color> getStones (){
        return stones;
    }

    @Column (name = "name")
    private String name = "Temple";

    @OneToOne
    @JoinColumn (name = "game_id")
    private Game game;

    @OneToOne
    @JoinColumn(name = "SHIP_ID")
    private Ship ship;

    public long getId(){
        return id;
    }

    public void fillDock (){isDockEmpty = false;}

    public void setId(long id) {
        this.id = id;
    }

    /*public void setStones(List<Stone> stones) {
        this.stones = stones;
    }*/

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public Game getGame() {
        return game;
    }

    public void setGame(Game game) {
        this.game = game;
    }

    public Ship getShip() {
        return ship;
    }

    public void setShip(Ship ship) {
        this.ship = ship;
    }
}

BoardService.java 类中的此处调用 (game.setTemple(newTemple)) 会引发错误:

   private void createAndAssignSites(Game game) {

        BurialChamber newBurialChamber = new BurialChamber();
        Pyramid newPyramid = new Pyramid();
        Obelisk newObelisk = new Obelisk();
        Temple newTemple = new Temple();
        Market newMarket = new Market();

        newBurialChamber.setGame(game);
        newPyramid.setGame(game);
        newObelisk.setGame(game);
        newTemple.setGame(game);
        newMarket.setGame(game);

//        game.setBurialChamber(newBurialChamber);
//        game.setPyramid(newPyramid);
//        game.setObelisk(newObelisk);
        game.setTemple(newTemple);
//        game.setMarket(newMarket);

        gameRepository.save(game);

        burialChamberRepository.save(newBurialChamber);
        pyramidRepository.save(newPyramid);
        obeliskRepository.save(newObelisk);
        templeRepository.save(newTemple);
        marketRepository.save(newMarket);

这里是没有 getter 和 setter 以及其他简单方法的 Game.java 类:

import java.io.Serializable;
import java.util.ArrayList;
import java.util.List;

import javax.persistence.*;


import ch.uzh.ifi.seal.soprafs17.constant.GameStatus;
import ch.uzh.ifi.seal.soprafs17.entity.sites.*;
import com.fasterxml.jackson.annotation.JsonIgnore;
import org.hibernate.annotations.LazyCollection;
import org.hibernate.annotations.LazyCollectionOption;

@Entity (name = "game")
public class Game implements Serializable {

    private static final long serialVersionUID = 1L;

    private List<User> players = new ArrayList<>();

    private List<Move> moves = new ArrayList<>();

    public Game (){}

    public Game (String name, long ownerID, User player){
        this.name = name;
        this.ownerID = ownerID;
        this.status = GameStatus.PENDING;
        players.add(player);
    }

    private Long id;

    @Id
    @GeneratedValue
    @Column (name = "game_id")
    public Long getId (){
        return id;
    }

    @Column(nullable = false)
    private String name;

    @Column(nullable = false) 
    private Long ownerID;

    @Column 
    private GameStatus status;

    @Column 
    private Integer currentPlayer = 0;

    @OneToMany(mappedBy="game")
    public List<Move> getMoves(){
        return moves;
    }

    @JsonIgnore
    @LazyCollection(LazyCollectionOption.FALSE)
    @OneToMany (mappedBy="game",cascade = CascadeType.ALL)
    public List<User> getPlayers (){
        return players;
    }

    public void setPlayers (List<User> players){
        this.players = players;
    }

    @OneToOne
    private BurialChamber burialChamber;

    @OneToOne
    private Market market;

    @OneToOne
    private Obelisk obelisk;

    @OneToOne (mappedBy = "game")
    private Pyramid pyramid;

    @OneToOne
    private Shipyard shipyard;

    @OneToOne (mappedBy = "game")
    private Temple temple;

我不明白那2722个字符串是什么以及它是在哪里生成的。以及为什么作业在一个方向有效，但在另一个方向却无效...... 希望大家能指出错误来源。

谢谢 阿里克

Answer 1

问题在于您在字段和方法上混合了注释。

JPA 提供程序将通过查找 @ID 注释(在 Game 的情况下位于方法上)来确定您使用的策略。

@Id
@GeneratedValue
@Column (name = "game_id")
public Long getId (){
    return id;
}

本质上，Temple 上的 @OneToOne 注释会被忽略，因为它在 Field 上:

@OneToOne (mappedBy = "game")
private Temple temple;

因此，Hibernate 实质上尝试将 Temple 作为二进制值保留在 Game 表中，因为它不知道这种关系 - 它只是将其视为一个简单的字段。

您可以按照下面的详细说明混合注释，但这很少需要。大多数情况下使用其中之一:

http://howtodoinjava.com/jpa/field-vs-property-vs-mixed-access-modes-jpa-tutorial/