我正在尝试使用 hibernate 映射一个实体，但是使用 SQL Server，我无法继续。
详情如下。
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 key
----------------------------
@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;
}
}
图书服务
--------------------
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");
}
}
问题是我不能使用序列并通过执行它 findBook 始终有效并持续失败并出现错误。
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。
最佳答案
让它工作的唯一方法是覆盖 SQLInsert 并欺骗期望设置标识符列的 Hibernate。如果您提供自己的自定义 INSERT 语句，则可以这样做，而不是将 rowId 设置为 null，而是设置版本:
@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;
}
}
完成此更改后，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 将生成正确的 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)]
测试可在 GitHub 上获得.
关于java - 当嵌入式键包含 SQL Server 上的标识列时，Hibernate 插入失败，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/46035660/