使用jpa和休眠问题的JoiningTables是mulitmapping失败的

我两个月前才刚刚开始学习Java ee，并且在某些方面都在挣扎中，如下所示。在完成一些逻辑后，我为预订系统提供了三个实体类，并且在运行项目时遇到了麻烦：

@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;

当我运行项目时，出现以下错误消息：

异常[EclipseLink-48]（Eclipse Persistence Services-2.5.2.v20140319-9ad6abd）：org.eclipse.persistence.exceptions.DescriptorException异常描述：[booking.SHIPID]字段存在多个可写映射。只能将其中一个定义为可写，所有其他都必须指定为只读。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：[booking.PASSID]字段存在多个可写映射。只能将其中一个定义为可写，所有其他都必须指定为只读。映射：org.eclipse.persistence.mappings.OneToOneMapping [passenger]描述符：

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

我了解到映射中存在问题，我对此进行了一些研究，但仍然无法了解如何解决它或如何建立实体类之间的关系。谁能帮助我找出问题并解决。

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;

问题来源：Stack Overflow