如何在Hibernate中将字符串映射到DB序列 [英] How to map a string to DB sequence in Hibernate

问题描述

标题中有很多这样说。我有一个看起来像这样的类：

  @Entity 
 @Table（name =FOO）
 public class Foo {
 
 private String theId; 
 
 @Id 
 @Column（name =FOO_ID）
 @GeneratedValue（strategy = GenerationType.SEQUENCE，generator =fooIdSeq）
 @SequenceGenerator（name =fooIdSeq，sequenceName =SQ_FOO_ID，allocationSize = 10）
 public String getTheId（）{return theId; } 
 
 public String setTheId（String theId）{this.theId = theId; } 
} 
  

使用Oracle 11g， FOO_ID 列是一个 VARCHAR2 ，但序列 SQ_FOO_ID 产生一个 NUMBER 。数据库显然对此感到满意，但应用程序需要能够支持在应用程序之外插入此列的非数字ID。

考虑到上面的代码中，我得到了一个 org.hibernate.id.IdentifierGenerationException：ids：java.lang.String 的未知整数数据类型。有没有什么办法做这种映射？

使用Hibernate 3.6。

解决方案

实现一个自定义的IdentifierGenerator类;从博文：

  import java.io.Serializable; 
 import java.sql.Connection; 
 import java.sql.PreparedStatement; 
 import java.sql.ResultSet; 
 import java.sql.SQLException; 
 
 import org.hibernate.HibernateException; 
 import org.hibernate.engine.spi.SessionImplementor; 
 import org.hibernate.id.IdentifierGenerator; 
 
 public class StringKeyGenerator implements IdentifierGenerator {
 $ b $ @Override 
 public Serializable generate（SessionImplementor session，Object collection）throws HibernateException {
 Connection connection = session。连接（）; 
 PreparedStatement ps = null; 
 String result =; 
 
 try {
 //特定于Oracle的代码来查询序列
 ps = connection.prepareStatement（SELECT TABLE_SEQ.nextval AS TABLE_PK FROM dual）; 
 ResultSet rs = ps.executeQuery（）; 
 
 if（rs.next（））{
 int pk = rs.getInt（TABLE_PK）; 
 
 //转换为字符串
 result = Integer.toString（pk）; 
} 
} catch（SQLException e）{
抛出新的HibernateException（Unable to generate Primary Key）; 
} finally {
 if（ps！= null）{
 try {
 ps.close（）; 
} catch（SQLException e）{
抛出新的HibernateException（Unable to close prepared statements。）; 
} 
} 
} 
 
返回结果; 
 
 
 
 
 
 
 $ b 
注释实体PK，如下所示：
  @Id 
 @GenericGenerator（name =seq_id，strategy =my.package.StringKeyGenerator）
 @ GeneratedValue（generator =seq_id）
 @Column（name =TABLE_PK，unique = true，nullable = false，length = 20）
 public String getId（）{
 return this。 ID; 
 
 
 
 $ b $ p 
 $ b 
由于Eclipse中存在一个错误，可能会引发一个错误，  seq_id ）未在持久性单元中定义。将此设置为如下警告： 
 
 
 
 
选择窗口»首选项 li>展开 Java持久性»JPA»错误/警告 
 
 
单击查询和生成器 >设置生成器未在持久性单元中定义为：警告 
 
 
单击 / strong>应用更改并关闭对话框
 
 
Pretty much says it in the title. I have a class that looks something like this:
@Entity
@Table(name="FOO")
public class Foo {

  private String theId;

  @Id
  @Column(name = "FOO_ID")
  @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "fooIdSeq")
  @SequenceGenerator(name = "fooIdSeq", sequenceName = "SQ_FOO_ID", allocationSize = 10)
  public String getTheId() { return theId; }

  public String setTheId(String theId) { this.theId = theId; }
}
Using Oracle 11g, the FOO_ID column is a VARCHAR2, but the sequence SQ_FOO_ID yields a NUMBER. The database is apparently happy with this, but the application needs to be able to support non-numeric IDs that may have been inserted into this column outside of the application.


Considering the code above, I get a org.hibernate.id.IdentifierGenerationException: Unknown integral data type for ids : java.lang.String. Is there any way to do this mapping?

Using Hibernate 3.6.
解决方案
Implement a custom IdentifierGenerator class; from a blog post:
import java.io.Serializable;
import java.sql.Connection;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;

import org.hibernate.HibernateException;
import org.hibernate.engine.spi.SessionImplementor;
import org.hibernate.id.IdentifierGenerator;

public class StringKeyGenerator implements IdentifierGenerator {

    @Override
    public Serializable generate(SessionImplementor session, Object collection) throws HibernateException {
        Connection connection = session.connection();
        PreparedStatement ps = null;
        String result = "";

        try {
            // Oracle-specific code to query a sequence
            ps = connection.prepareStatement("SELECT TABLE_SEQ.nextval AS TABLE_PK FROM dual");
            ResultSet rs = ps.executeQuery();

            if (rs.next()) {
                int pk = rs.getInt("TABLE_PK");

                // Convert to a String
                result = Integer.toString(pk);
            }
        } catch (SQLException e) {
            throw new HibernateException("Unable to generate Primary Key");
        } finally {
            if (ps != null) {
                try {
                    ps.close();
                } catch (SQLException e) {
                    throw new HibernateException("Unable to close prepared statement.");
                }
            }
        }

        return result;
    }
}
Annotate the entity PK like this:
@Id
@GenericGenerator(name="seq_id", strategy="my.package.StringKeyGenerator")
@GeneratedValue(generator="seq_id")
@Column(name = "TABLE_PK", unique = true, nullable = false, length = 20)
public String getId() {
    return this.id;
}
Due to a bug in Eclipse, an error might be raised that the generator (seq_id) is not defined in the persistence unit. Set this to a warning as follows:

Select Window » Preferences
Expand Java Persistence » JPA » Errors/Warnings
Click Queries and generators
Set Generator is not defined in the persistence unit to: Warning
Click OK to apply changes and close the dialog


                        
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