使用 Hibernate 注释映射 PostgreSQL 串行类型
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Mapping PostgreSQL serial type with Hibernate annotations
提问by alecswan
I am using Hibernate 3.3 and PostgreSQL 8.x and would like to use Hibernate annotations to map an auto-incremented column which is NOT a primary key.
我正在使用 Hibernate 3.3 和 PostgreSQL 8.x，并想使用 Hibernate 注释来映射一个不是主键的自动递增列。
It doesn't matter if the column is mapped using SERIAL type or sequences in Postgres as long as it gets auto-incremented by the database and not by Hibernate. I tried the following mappings, but they always generated null orderId.
列是否使用 SERIAL 类型或 Postgres 中的序列映射并不重要，只要它由数据库而不是 Hibernate 自动递增即可。我尝试了以下映射，但它们总是生成空 orderId。
@Column(name = "orderId", insertable = false)
@Generated(GenerationTime.INSERT)
//@GeneratedValue(strategy = javax.persistence.GenerationType.AUTO)
private Integer orderId;
I will appreciate any help with this.
我将不胜感激。
Thanks
谢谢
回答by axtavt
The following mapping should work fine:
以下映射应该可以正常工作：
@Column(name = "orderId")
@Generated(GenerationTime.INSERT)
private Integer orderId;
Note, however, that generated value for freshly saved objects is not available until session is flushed.
但是请注意，在刷新会话之前，新保存的对象的生成值不可用。
EDIT:Note that this mapping doesn't affect doesn't make Hibernate to create a column of type serialduring schema generation, since Hibernate doesn't know anything about the nature of value generation at the database side. Therefore, if you want Hibernate to create a column with a proper type, you need to specifiy it explicitly:
编辑：请注意，此映射不会影响不会使 Hibernateserial在模式生成期间创建类型的列，因为 Hibernate 对数据库端的值生成的性质一无所知。因此，如果您希望 Hibernate 创建具有适当类型的列，则需要明确指定它：
@Column(name = "orderId", columnDefinition = "serial")
@Generated(GenerationTime.INSERT)
private Integer orderId;
And on a recent Hibernate version (4.3), you can use this:
在最近的 Hibernate 版本 (4.3) 上，您可以使用以下命令：
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Long orderId;
回答by pstanton
the accepted answer doesn't work for me.
接受的答案对我不起作用。
this did though:
虽然这样做了：
@Id
@Column(name = "your_id", columnDefinition = "serial")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer yourId;
回答by bjonczy
I'm using this with postgresql9.1, should work with 8 too:
我将它与 postgresql9.1 一起使用，也应该与 8 一起使用：
@SequenceGenerator(allocationSize=1, initialValue=1, sequenceName="account_id_seq", name="account_id_seq")
@GeneratedValue(generator="account_id_seq", strategy=GenerationType.SEQUENCE)
@Id
@Column(name="id")
private Integer id;
