当嵌入式键包含 SQL Server 上的标识列时，休眠插入失败

Posted 2023-03-29

【中文标题】当嵌入式键包含 SQL Server 上的标识列时，休眠插入失败

【英文标题】：Hibernate insert failing when embedded key contains identity column on SQL Server

【发布时间】：2018-02-12 14:37:54

【问题描述】：

我正在尝试使用 hibernate 映射实体，但使用 SQL Server，我无法继续。

以下是详细信息。

SQL Server 实体

CREATE TABLE [dbo].[BOOK_EMBEDDED](

[row_id] [bigint] IDENTITY(1,1) NOT NULL,

[group_no] [int] NOT NULL,

[book_name] [varchar](255) NULL,

 CONSTRAINT [PK_BOOK_EMBEDDED] PRIMARY KEY CLUSTERED 

(

[group_no] ASC,

[row_id] ASC

)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]

) ON [PRIMARY]

==============================

嵌入的密钥

---------------

@Embeddable 

public class EmbeddedKey implements Serializable  


    private static final long serialVersionUID = 1L; 



    @GeneratedValue(strategy = GenerationType.IDENTITY) 

    @Column(name = "row_id") 

    private Long rowId; 



    @Column(name = "group_no") 

    private int groupNo; 



    public Long getRowId()  

        return rowId; 

     



    public void setRowId(Long rowId)  

        this.rowId = rowId; 

     



    public static long getSerialversionuid()  

        return serialVersionUID; 

     



    @Override 

    public int hashCode()  

        final int prime = 31; 

        int result = 1; 

        result = (int) (prime * result + rowId); 

        result = prime * result + groupNo; 

        return result; 

     



    @Override 

    public boolean equals(Object obj)  

        if (this == obj) 

            return true; 

        if (obj == null) 

            return false; 

        if (getClass() != obj.getClass()) 

            return false; 

        EmbeddedKey other = (EmbeddedKey) obj; 

        if (rowId != other.rowId) 

            return false; 

        if (groupNo != other.groupNo) 

            return false; 

        return true; 

     



    @Override 

    public String toString()  

        return this.getRowId() + "  " + this.getGroupNo() + " "; 

     



    public int getGroupNo()  

        return groupNo; 

     



    public void setGroupNo(int groupNo)  

        this.groupNo = groupNo; 

     



 

实体

-------------

@Entity(name = "BOOK_EMBEDDED") 

public class Bookmysql implements Serializable  



    /** 

     *  

     */ 

    private static final long serialVersionUID = 1L; 



    @Column(name = "BOOK_NAME") 

    private String book_Name; 



    @EmbeddedId 

    private EmbeddedKey key; 



    public BookMySQL()  

     



    public String getBook_Name()  

        return book_Name; 

     



    public void setBook_Name(String book_Name)  

        this.book_Name = book_Name; 

     



    public static long getSerialversionuid()  

        return serialVersionUID; 

     



    public EmbeddedKey getKey()  

        return key; 

     



    public void setKey(EmbeddedKey key)  

        this.key = key; 

     



    @Override 

    public String toString()  

        return this.getKey().toString() + "  " + this.getBook_Name(); 

     



 

实体管理器类

------------------

public class LocalEntityManager  

    private static EntityManagerFactory emf; 

    private static EntityManager em; 



    private LocalEntityManager()  

     



    public static EntityManager getEntityManger()  

        if (emf == null)  

            synchronized (LocalEntityManager.class)  

                if (emf == null)  

                    emf = Persistence.createEntityManagerFactory("BookEntities"); 

                    em = emf.createEntityManager(); 

                 

             

         

        return em; 

     

 

预订服务

------------------

public class MySQLBookService  



    public Long persistBook(String bookName)  

        EmbeddedKey key = new EmbeddedKey(); 

        key.setGroupNo(1); 



        BookMySQL book = new BookMySQL(); 

        book.setBook_Name(bookName); 

        book.setKey(key); 



        EntityManager em = LocalEntityManager.getEntityManger(); 

        EntityTransaction tx = em.getTransaction(); 

        tx.begin(); 

        em.persist(book); 

        tx.commit(); 

        em.close(); 



        return book.getKey().getRowId(); 

     



    public BookMySQL findBook(int bookId)  

        EntityManager em = LocalEntityManager.getEntityManger(); 

        EmbeddedKey key = new EmbeddedKey(); 

        key.setGroupNo(1); 

        key.setRowId(1L); 

        BookMySQL bookMySQL = em.find(BookMySQL.class, key); 

        System.out.println(bookMySQL); 

        return bookMySQL; 

     



    public static void main(String... args)  

        MySQLBookService bookService = new MySQLBookService(); 

        // bookService.findBook(1); 



        bookService.persistBook("Lord of the rings"); 

     



 

问题是我不能使用一个序列并通过执行这个 findBook 始终有效并且持久失败并出现错误。

ERROR: Cannot insert explicit value for identity column in table 'BOOK_EMBEDDED' when IDENTITY_INSERT is set to OFF.

任何帮助将不胜感激。

【问题讨论】：

Yoy可以按照SO规则格式化代码吗？

【参考方案1】：

使其工作的唯一方法是覆盖SQLInsert 并欺骗希望设置标识符列的Hibernate。如果您提供自己的自定义 INSERT 语句，则可以做到这一点，这样您就可以设置版本，而不是将 rowId 设置为 null：

@Entity(name = "BOOK_EMBEDDED")
@SQLInsert( sql = "insert into BOOK_EMBEDDED (BOOK_NAME, group_no, version) values (?, ?, ?)")
public static class Book implements Serializable 

    @EmbeddedId
    private EmbeddedKey key;

    @Column(name = "BOOK_NAME")
    private String bookName;

    @Version
    @Column(insertable = false)
    private Integer version;

    public EmbeddedKey getKey() 
        return key;
    

    public void setKey(EmbeddedKey key) 
        this.key = key;
    

    public String getBookName() 
        return bookName;
    

    public void setBookName(String bookName) 
        this.bookName = bookName;
    

进行此更改后，您可以运行以下测试：

doInJPA(entityManager -> 

    EmbeddedKey key = new EmbeddedKey();
    key.setGroupNo(1);

    Book book = new Book();
    book.setBookName( "High-Performance Java Persistence");

    book.setKey(key);

    entityManager.persist(book);
);

doInJPA(entityManager -> 
    EmbeddedKey key = new EmbeddedKey();

    key.setGroupNo(1);
    key.setRowId(1L);

    Book book = entityManager.find(Book.class, key);
    assertEquals( "High-Performance Java Persistence", book.getBookName() );
);

Hibernate 会生成正确的 SQL 语句：

Query:["insert into BOOK_EMBEDDED (BOOK_NAME, group_no, version) values (?, ?, ?)"], Params:[(High-Performance Java Persistence, 1, NULL(BIGINT))]

Query:["select compositei0_.group_no as group_no1_0_0_, compositei0_.row_id as row_id2_0_0_, compositei0_.BOOK_NAME as BOOK_NAM3_0_0_, compositei0_.version as version4_0_0_ from BOOK_EMBEDDED compositei0_ where compositei0_.group_no=? and compositei0_.row_id=?"], Params:[(1, 1)]

测试可通过GitHub 获得。

【讨论】：

有趣，但实际上hibernate论坛（forum.hibernate.org/viewtopic.php?f=1&t=1006818）建议做不到。它是否适用于venilla hibernate，我只尝试过hibernate（无JPA）但无法实现..如果你愿意，我可以发布错误

那个论坛的答案是 2010 年的，我的是 2017 年的，并且有一个运行良好的测试用例，您可以在 GitHub 上找到。只需 fork 我的高性能 Java 持久性图书存储库并运行测试。

是的。当然，我确实调查过。我会尝试并更新你。同样，我可以期望它与 vanilla(no jpa) hibernate 一起工作吗？

您也可以使该存储库中的所有测试都以休眠模式运行。查看 AbstractTest 并检查本机引导程序。

当然@Vlad Mihalcea，谢谢..我今天会尝试锻炼。

【参考方案2】：

您必须将 identity_insert 设置为 ON 才能使其工作。

查看这篇文章 - How to set IDENTITY_INSERT

所以运行下面的命令：

SET IDENTITY_INSERT BOOK_EMBEDDED ON

【参考方案3】：

只需删除和移除 EmbeddedKey 类并将字段添加到您的实体。

import java.io.*;
import javax.persistence.*;

@Entity(name = "Book")
@Table(name = "book")
public class Book implements Serializable


  @Id
  @Basic(optional = false)
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  @Column(name = "registration_number")
  private Long registrationNumber;

  @Column(name = "publisher_id")
  private Integer publisherId;

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

  //Getters and setters omitted for brevity

CREATE TABLE book (
        publisher_id INT NOT NULL,
        registration_number BIGINT IDENTITY NOT NULL,
        title VARCHAR(255),
        PRIMARY KEY (publisher_id, registration_number)
    )