當嵌入的鍵包含 SQL Server 上的標識列時，休眠插入失敗

Question

我正在嘗試使用休眠映射實體，但是使用 SQL Server，我無法繼續。

以下是詳細信息。

SQL Server 實體

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]

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

嵌入式密鑰

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

@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; 

    } 



} 

實體

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

@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(); 

    } 



} 

實體管理器類

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

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; 

    } 

} 

訂書服務

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

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"); 

    } 



} 

問題是我無法使用序列，並且通過執行此 findBook 始終有效，並且持久失敗並出現錯誤。

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

任何幫助將不勝感激。

Answer 1

使其工作的唯一方法是覆蓋SQLInsert並欺騙期望設置標識符列的 Hibernate。 如果您提供自己的自定義 INSERT 語句，則可以完成此操作，以便將rowId設置為 null，而不是將版本設置為：

@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;
    }
}

進行此更改后，您可以運行以下測試：

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() );
});

並且 Hibernate 會生成正確的 SQL 語句：

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)]

該測試可在GitHub 上獲得。

Answer 2

您必須將 identity_insert 設置為 ON 才能使其工作。

檢查這篇文章 -如何設置 IDENTITY_INSERT

所以運行下面的命令：

SET IDENTITY_INSERT BOOK_EMBEDDED ON

Answer 3

只需刪除並移除 EmbeddedKey 類並將字段添加到您的實體中即可。

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)
    )