JoiningTables 使用 jpa 和休眠问题是多映射失败

Question

两个月前我刚开始学习 java ee，我在某些方面很挣扎，如下所示。 我有三个用于预订系统的实体类，在完成一些逻辑后，我在努力运行项目：

     @Entity
     @Table(name="booking")
     public class Booking implements Serializable {
     @Id
     @GeneratedValue(strategy = GenerationType.IDENTITY)
     private int reservationId;
     private String stateroomType;
     private double totalAmount;
     private int totalGuests;   
     private int shipId;
     private int passId;

  //Joining Tables
  @OneToOne
  @JoinColumn(name="passId")
  private Passenger passenger;

  @ManyToOne
  @JoinColumn(name="shipId")
  private Cruise cruise;

    @Entity
     @Table(name = "shipcruise")
    public class Cruise implements Serializable {


    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int cruiseId;
    private String cruiseName;
    private LocalDate startDate;
    private LocalDate endDate;
    private Timestamp destination;

    @Entity
    @Table(name = "passengers")
    public class Passenger implements Serializable{

    @Id
   @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int passengerId;
    private String userName;
    private String firstname;
    private String lastname;
    private String address;
    private String city;
    private String country;
    private String postalCode;
    private String password;

当我运行我的项目时，我收到以下错误消息：

异常 [EclipseLink-48] (Eclipse Persistence Services - 2.5.2.v20140319-9ad6abd): org.eclipse.persistence.exceptions.DescriptorException 异常描述：字段 [booking.SHIPID] 存在多个可写映射。只有一个可以定义为可写，所有其他必须指定为只读。 映射：org.eclipse.persistence.mappings.OneToOneMapping[cruise] 描述符：RelationalDescriptor(com.springmvc.jpa.booking.Booking --> [DatabaseTable(booking)]) 异常 [EclipseLink-48] (Eclipse Persistence Services - 2.5.2.v20140319-9ad6abd): org.eclipse.persistence.exceptions.DescriptorException 异常描述：字段 [booking.PASSID] 存在多个可写映射。只有一个可以定义为可写，所有其他必须指定为只读。 映射：org.eclipse.persistence.mappings.OneToOneMapping[passenger] 描述符：RelationalDescriptor(com.springmvc.jpa.booking.Booking --> [DatabaseTable(booking)])

.................................................. .....................................................

我了解映射存在问题，我对此进行了一些研究，但仍然不知道如何解决它或如何建立实体类之间的关系。谁能帮我找出问题并解决它。

数据库表：

      CREATE TABLE `booking` (
     `reservationId` int NOT NULL,
     `stateroomType` varchar(30) NOT NULL,
     `totalGuests` int NOT NULL,
     `totalAmount` decimal(10,2) NOT NULL,
     `passId` int DEFAULT NULL,
     `shipId` int DEFAULT NULL,
      PRIMARY KEY (`reservationId`),
      KEY `passId` (`passId`),
      KEY `shipId` (`shipId`),
      CONSTRAINT `booking_ibfk_1` FOREIGN KEY (`passId`) REFERENCES 
      `passengers` (`passengerId`),
      CONSTRAINT `booking_ibfk_2` FOREIGN KEY (`shipId`) REFERENCES 
      `shipcruise` (`cruiseId`)
       ) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci;

      CREATE TABLE `passengers` (
     `passengerId` int NOT NULL AUTO_INCREMENT,
     `userName` varchar(50) DEFAULT NULL,
     `password` varchar(25) DEFAULT NULL,
     `firstname` varchar(30) DEFAULT NULL,
     `lastname` varchar(30) DEFAULT NULL,
     `address` varchar(255) DEFAULT NULL,
    `city` varchar(25) DEFAULT NULL,
    `postalCode` varchar(10) DEFAULT NULL,
     `country` varchar(20) DEFAULT NULL,
     PRIMARY KEY (`passengerId`)
     ) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=utf8mb4 
   COLLATE=utf8mb4_0900_ai_ci;

    CREATE TABLE `shipcruise` (
   `cruiseId` int NOT NULL AUTO_INCREMENT,
   `CruiseName` varchar(50) DEFAULT NULL,
   `shipName` varchar(50) DEFAULT NULL,
  `startDate` date NOT NULL,
  `endDate` date NOT NULL,
   `destination` timestamp NOT NULL,
   PRIMARY KEY (`cruiseId`)
  ) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci;

Answer 1

预订实体。

 @Entity
 @Table(name="booking")
 public class Booking implements Serializable {
 @Id
 @GeneratedValue(strategy = GenerationType.IDENTITY)
 private Integer reservationId;
 private String stateroomType;
 private double totalAmount;
 private int totalGuests;   

 //Joining Tables
 @ManyToOne
 @JoinColumn(name="passId")
 private Passenger passenger;

 @ManyToOne
 @JoinColumn(name="shipId")
 private Cruise cruise;

巡航实体。

@Entity
@Table(name = "shipcruise")
public class Cruise implements Serializable {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer cruiseId;
private String cruiseName;
private LocalDate startDate;
private LocalDate endDate;
private Timestamp destination;

乘客实体。

@Entity
@Table(name = "passengers")
public class Passenger implements Serializable{

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer passengerId;
private String userName;
private String firstname;

在你的代码中，你有：

 private int shipId;
 private int passId;

以下，你有：

  //Joining Tables
  @OneToOne
  @JoinColumn(name="passId")
  private Passenger passenger;

  @ManyToOne
  @JoinColumn(name="shipId")
  private Cruise cruise;

当您不使用@Column 或@JoinColumn 时，eclipse-link 将使用字段名称，因此，在这种情况下，您将拥有 2 个指向同一列的 java 属性。

@JoinColumn 为您完成了繁琐的工作（引用引用 SQL 表的其他实体），这就是我们使用 JPA 的原因。

我将表示主键的字段从 int 更改为 Integer 类。 你可以在这里找到原因：What's difference between primitive and wrapper class in JPA (Hibernate) column mappings?