使用 JPA 持久保存 1 列表

发布于 2024-11-24 07:31:55 字数 1777 浏览 0 评论 0 原文

我有一张只有一列的表格。这是一个 Id

我怎样才能用 JPA 持久化它？

我尝试过 entityManager.persist(new OneColumnTable());

并且它抛出 PersistenceException

Caused by: javax.persistence.PersistenceException: Exception [EclipseLink-6023] (Eclipse Persistence Services - 2.2.0.v20110202-r8913): org.eclipse.persistence.exceptions.QueryException
Exception Description: **The list of fields to insert into the table [DatabaseTable(OneColumnTable)] is empty.  You must define at least one mapping for this table.**
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.flush(EntityManagerImpl.java:747)
    at com.sun.enterprise.container.common.impl.EntityManagerWrapper.flush(EntityManagerWrapper.java:418)

我该怎么办？

更新

@Entity
@Table(name = "OneColumnTable")
public class OneColumnTable implements Serializable{
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    @Column(name = "OneColumn")
    private Integer oneColumn;

    public OneColumnTable() {

    }

    public Integer getOneColumn() {
        return oneColumn;
    }

    public void setOneColumn(Integer oneColumn) {
        this.oneColumn= oneColumn;
    }

}

表格

USE [myDB]
GO
/****** Object:  Table [dbo].[OneColumnTable]    Script Date: 07/15/2011 12:10:56 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TABLE [dbo].[OneColumnTable](
    [OneColumn] [bigint] IDENTITY(1,1) NOT NULL,
PRIMARY KEY CLUSTERED 
(
    [OneColumn] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]

原文

I've got a table with only one column. It's an Id

How can I persist it with JPA?

I've tried a entityManager.persist(new OneColumnTable());

and it throws a PersistenceException

Caused by: javax.persistence.PersistenceException: Exception [EclipseLink-6023] (Eclipse Persistence Services - 2.2.0.v20110202-r8913): org.eclipse.persistence.exceptions.QueryException
Exception Description: **The list of fields to insert into the table [DatabaseTable(OneColumnTable)] is empty.  You must define at least one mapping for this table.**
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.flush(EntityManagerImpl.java:747)
    at com.sun.enterprise.container.common.impl.EntityManagerWrapper.flush(EntityManagerWrapper.java:418)

How can I do it?

UPDATE

@Entity
@Table(name = "OneColumnTable")
public class OneColumnTable implements Serializable{
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    @Column(name = "OneColumn")
    private Integer oneColumn;

    public OneColumnTable() {

    }

    public Integer getOneColumn() {
        return oneColumn;
    }

    public void setOneColumn(Integer oneColumn) {
        this.oneColumn= oneColumn;
    }

}

TABLE

USE [myDB]
GO
/****** Object:  Table [dbo].[OneColumnTable]    Script Date: 07/15/2011 12:10:56 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TABLE [dbo].[OneColumnTable](
    [OneColumn] [bigint] IDENTITY(1,1) NOT NULL,
PRIMARY KEY CLUSTERED 
(
    [OneColumn] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]

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花之痕靓丽 2024-12-01 07:31:55

DataNucleus 可以很好地保留该类。问题是，对于某些 RDBMS，当未指定列时，INSERT 语句必须是特定类型（因为您唯一的列是在数据存储中生成的），而这可能是它无法做到的。对于 SQLServer，任何合适的 JPA 实现都应该生成的语句是“INSERT INTO {tbl} DEFAULT VALUES”。也许得到一个可以做到这一点的实现？

回复收藏 0 原文

云淡风轻 2024-12-01 07:31:55

@GenerateValue(strategy=GenerationType.IDENTITY) 指示使用依赖于对 IDENTITY 列的数据库支持。

IDENTITY 列的概念并不存在于所有数据库中。例如，Apache Derby 和 Oracle 不支持这一点，而 MySQL、MSSQL（通常是 Sybase）则支持 IDENTITY 排序策略。您应该使用数据库支持的排序策略。为了可移植性，请选择 AUTO 或 TABLE 排序策略。在大多数 JPA 提供程序中，AUTO 策略被实现为 TABLE 排序策略，因为所有数据库都支持创建用于维护序列值的表。