java 休眠自动增量不起作用

Hibernate Auto-Increment not working

提问by dharga

I have a column in my DB that is set with Identity(1,1) and I can't get hibernate annotations to work for it. I get errors when I try to create a new record.

我的数据库中有一个列设置了 Identity(1,1) 并且我无法获得休眠注释来为它工作。尝试创建新记录时出现错误。

In my entity I have the following.

在我的实体中，我有以下内容。

@Entity
@Table(schema="dbo", name="MemberSelectedOptions")
public class MemberSelectedOption extends BampiEntity implements Serializable {

    @Embeddable
    public static class MSOPK implements Serializable {
        private static final long serialVersionUID = 1L;

        @Column(name="SourceApplication")
        String sourceApplication;

        @Column(name="GroupId")
        String groupId;

        @Column(name="MemberId")
        String memberId;

        @Column(name="OptionId")
        int optionId;

        @GeneratedValue(strategy=GenerationType.IDENTITY, generator="native")
        @Column(name="SeqNo", unique=true, nullable=false)
        BigDecimal seqNo;

        //Getters and setters here...

    }

    private static final long serialVersionUID = 1L;

    @EmbeddedId
    MSOPK pk = new MSOPK();

    @Column(name="OptionStatusCd")
    String optionStatusCd;

    @Column(name="EffectiveDate")
    Date effectiveDate;

    @Column(name="TermDate")
    Date termDate;

    @Column(name="SelectionStatusDate")
    Date selectionStatusDate;   

    @Column(name="SysLstUpdtUserId")
    String sysLstUpdtUserId = Globals.WS_USER_ID;;

    @Column(name="SysLstTrxDtm")
    Date sysLstTrxDtm = new Date();

    @OneToMany(mappedBy="option")
    List<MemberSelectedVariable> variables = 
                             new ArrayList<MemberSelectedVariable>();

        //More Getters and setters here...
}

But when I try to add a new record I get the following error.

但是当我尝试添加新记录时，出现以下错误。

Cannot insert explicit value for identity column in table 'MemberSelectedOptions' when IDENTITY_INSERT is set to OFF. I don't want to set IDENTIY_INSERT to ON because I want the identity column in the db to manage the values.

当 IDENTITY_INSERT 设置为 OFF 时，无法为表“MemberSelectedOptions”中的标识列插入显式值。我不想将 IDENTIY_INSERT 设置为 ON，因为我希望数据库中的标识列管理这些值。

The SQL that is run is the following; where you can clearly see the insert.

运行的 SQL 如下；在那里您可以清楚地看到插入物。

insert into dbo.MemberSelectedOptions 
  (OptionStatusCd, 
  EffectiveDate,
  TermDate, 
  SelectionStatusDate, 
  SysLstUpdtUserId, 
  SysLstTrxDtm, 
  SourceApplication,
  GroupId,
  MemberId, 
  OptionId, 
  SeqNo) 
values (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)

What am I missing?

我错过了什么？

回答by Andrey

this combination works great for me:

这种组合对我很有用：

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)

回答by Pascal Thivent

When you use @Embeddableor @EmbeddedId, the primary key values are supposed to be providedby the application (i.e. made up of non generated values). Your @GeneratedValueannotation is just ignored.

当您使用@Embeddableor 时@EmbeddedId，主键值应该由应用程序提供（即由非生成值组成）。您的@GeneratedValue注释只是被忽略。

回答by Rupeshit

Here is the example to do it

这是执行此操作的示例

@Id
@Column(name = "col_id")
@GeneratedValue(strategy = GenerationType.AUTO)
private Long             colId;

回答by baklarz2048

You can't do it with Create table manually and everything will be ok.

你不能用手动创建表来做，一切都会好的。

CREATE TABLE `Forum` (
  `name` varchar(255) NOT NULL,
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `body` varchar(500) DEFAULT NULL,
  PRIMARY KEY (name,`id`),
  UNIQUE KEY `id` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin2




import java.io.Serializable;

import javax.persistence.Column;
import javax.persistence.EmbeddedId;
import javax.persistence.Entity;

@Entity
public class Forum implements Serializable {

    @EmbeddedId
    private ForumCompositePK forumPK;
    /**
     * 
     */
    private static final long serialVersionUID = 7070007885798411858L;

    @Column(length = 500)
    String body;

    public String getBody() {
        return body;
    }

    public void setBody(String body) {
        this.body = body;
    }

    public void setForumPK(ForumCompositePK forumPK) {
        this.forumPK = forumPK;
    }

    public  ForumCompositePK getForumPK() {
        return forumPK;
    }

}




import java.io.Serializable;

import javax.persistence.Column;
import javax.persistence.Embeddable;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;

@Embeddable
public class ForumCompositePK implements Serializable {

    /**
     * 
     */
    private static final long serialVersionUID = 8277531190469885913L;


    @Column(unique=true,updatable=false,insertable=false)
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    private Integer id;

    private String name;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }   

    public Integer getId() {
        return id;
    }   

    public void setId(Integer id) {
        this.id = id;
    }



}

回答by dharga

You can't use Generators on composite keys

您不能在复合键上使用生成器

回答by Serge S.

Possible you need to mark your field with @idand not specify generatorproperty.

您可能需要使用@id而不是指定generator属性来标记您的字段。

As showed in Hibernate Annotation - 2.2.3.1. Generating the identifier property, the next example uses the identity generator:

如Hibernate Annotation - 2.2.3.1 所示。生成标识符属性，下一个示例使用身份生成器：

@Id @GeneratedValue(strategy=GenerationType.IDENTITY)
public Long getId() { ... }