Question

在标题中几乎说出来了。我有一个看起来像这样的课程：

@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列为VARCHAR2，但序列SQ_FOO_ID产生NUMBER。数据库显然对此感到满意，但应用程序需要能够支持可能已在应用程序之外插入此列的非数字ID。

考虑到上面的代码，我得到了org.hibernate.id.IdentifierGenerationException: Unknown integral data type for ids : java.lang.String。有没有办法做这个映射？

使用Hibernate 3.6。

Answer 1

实现自定义IdentifierGenerator类;来自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;
    }
}

像这样注释实体PK：

@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;
}

由于Eclipse中存在错误，可能会引发错误，即生成器（seq_id）未在持久性单元中定义。将此设置为警告，如下所示：

选择窗口»偏好设置
展开 Java持久性»JPA»错误/警告
点击查询和生成器
将生成器未在持久性单元中设置为：Warning
点击确定以应用更改并关闭对话框

Answer 2

这是另一种方法：

import java.io.Serializable;

import org.hibernate.engine.spi.SessionImplementor;
import org.hibernate.id.IdentifierGeneratorHelper.BigDecimalHolder;
import org.hibernate.id.IntegralDataTypeHolder;
import org.hibernate.id.SequenceGenerator;

public class StringSequenceGenerator extends SequenceGenerator {
    @Override
    public Serializable generate(SessionImplementor session, Object obj) {
        return super.generate( session, obj ).toString();
    }

    protected IntegralDataTypeHolder buildHolder() {
        return new BigDecimalHolder();
    }
}

必须在id属性上指定sequence参数，如下例所示：

@Id
@GenericGenerator(name = "STRING_SEQUENCE_GENERATOR", strategy = "mypackage.StringSequenceGenerator", parameters = { @Parameter(name = "sequence", value = "MY_SEQUENCE_NAME") })
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "STRING_SEQUENCE_GENERATOR")
@Column(name = "MY_ID")
public String getMyId() {
    return this.myId;
}

如何在Hibernate中将字符串映射到DB序列

2 个答案: