Java 使用jpa和hibernate连接表的问题是多重映射失败
我两个月前才开始学习JavaEE,我在某些方面遇到了困难,如下所示。 我有三个实体类用于预订系统,在我完成了一些逻辑之后,我在运行项目方面遇到了困难:Java 使用jpa和hibernate连接表的问题是多重映射失败,java,mysql,spring-mvc,jpa,Java,Mysql,Spring Mvc,Jpa,我两个月前才开始学习JavaEE,我在某些方面遇到了困难,如下所示。 我有三个实体类用于预订系统,在我完成了一些逻辑之后,我在运行项目方面遇到了困难: @Entity @Table(name="booking") public class Booking implements Serializable { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) private in
@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;
运行项目时,我收到以下错误消息:
异常[EclipseLink-48](Eclipse持久性服务-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持久性服务-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)])
徖
我知道映射中存在一个问题,我对此进行了一些研究,但仍然无法了解如何解决该问题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;
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;
邮轮实体
@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;
在您的代码中,您有:
private int shipId;
private int passId;
在下面,您有:
//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链接将使用字段名,因此,在本例中,您将有两个指向同一列的java属性
@JoinColumn为您做了大量的工作(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类更改为Integer类。
您可以在这里找到原因: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;
邮轮实体
@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;
在您的代码中,您有:
private int shipId;
private int passId;
在下面,您有:
//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链接将使用字段名,因此,在本例中,您将有两个指向同一列的java属性
@JoinColumn为您做了大量的工作(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类更改为Integer类。
您可以在这里找到原因:您好,欢迎来到StackOverflow。在实体预订中,您定义了两个字段,它们指向数据库中的同一列。私人国际船舶识别号;和@JoinColumn(name=“passId”)私人乘客;如果不指定列的名称,JPA将使用属性名。感谢您的回复。我对这些关系感到困惑。你能举例说明一下吗。我将数据库中的shipId和passId加入到预订表中。我的坏消息。我复制并粘贴了不正确的属性。您拥有Manytone JoinColumn(name=“shipId”)私人游轮;这与:private int shipId相同;我试图以这种方式加入,但它给出了一个错误,因此我必须在booking实体类中声明shId和passId。我收到以下错误消息:::无法在表“booking”上解析加入列“shipId”,欢迎使用StackOverflow。在实体预订中,您定义了两个字段,它们指向数据库中的同一列。私人国际船舶识别号;和@JoinColumn(name=“passId”)私人乘客;如果不指定列的名称,JPA将使用属性名。感谢您的回复。我对这些关系感到困惑。你能举例说明一下吗。我将数据库中的shipId和passId加入到预订表中。我的坏消息。我复制并粘贴了不正确的属性。您拥有Manytone JoinColumn(name=“shipId”)私人游轮;这与:private int shipId相同;我试图以这种方式加入,但出现了一个错误,因此我必须在booking实体类中声明shId和passId。我收到以下错误消息:::无法在表“booking”上解析加入列“shipId”。我感谢您的努力。谢谢你的努力。谢谢