[英]Simple JPA to Postgresql 9.1 HibernateEx
似乎很簡單。 我過去在MySQL和Oracle上使用過Hibernate序列,但現在在Postgresql 9.1上使用JPA2 / Hibernate4,並不斷獲取異常。 有任何想法嗎? 我嘗試了不同類型的GenerationType.SEQUENCE和GenerationType.AUTO,但是沒有任何效果。 我絕對需要幫助。
[1] Postgresql 9.1 DDL
-- removed CACHE 100
create sequence my_seq INCREMENT 1 MINVALUE 1 MAXVALUE 2147483647 START 1;
create table my_table (
id integer NOT NULL DEFAULT nextval('my_seq'),
ident text NOT NULL CONSTRAINT ident_uniq UNIQUE,
cname text NOT NULL,
created timestamp WITH TIME ZONE NOT NULL DEFAULT ('now'::text)::timestamp(6) with time zone,
lastmodified timestamp WITH TIME ZONE NOT NULL DEFAULT ('now'::text)::timestamp(6) with time zone,
CONSTRAINT my_table_pkey PRIMARY KEY(id)
);
insert into my_table (ident, cname) values ('test001', 'Test 001');
[2] Persistence.xml
<?xml version="1.0" encoding="UTF-8"?>
<persistence xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"
version="2.0" >
<persistence-unit name="MyPersistenceUnit" transaction-type="RESOURCE_LOCAL">
<description>Persistence unit</description>
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<properties>
<property name="hibernate.dialect" value="org.hibernate.dialect.PostgreSQLDialect"/>
<property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver" />
<property name="javax.persistence.jdbc.url" value="jdbc:postgresql://localhost:5432/mydb" />
<property name="javax.persistence.jdbc.user" value="user" />
<property name="javax.persistence.jdbc.password" value="password" />
<property name="hibernate.hbm2ddl.auto" value="validate"/>
<property name="hibernate.show_sql" value="false" />
<property name="hibernate.format_sql" value="false" />
<property name="prefer_sequence_per_entity" value="true" />
</properties>
</persistence-unit>
</persistence>
[3]實體代碼
package com.mypackage;
import java.sql.Timestamp;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.SequenceGenerator;
import javax.persistence.Transient;
@Entity(name="my_table")
public class MyTable {
@SequenceGenerator(name="foo", sequenceName="my_seq", allocationSize=1, initialValue=1)
@GeneratedValue(strategy = GenerationType.IDENTITY, generator = "foo")
@Id
@Column(name = "id")
private long id;
private String ident;
private Timestamp created;
private Timestamp lastmodified;
}
[5] JUnit
public class Test_MyTable {
@Test
public void test_1() {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("MyPersistenceUnit");
EntityManager em = emf.createEntityManager();
List<MyTable> list = em.createQuery("from my_table", MyTable.class).getResultList();
int size = ( (list == null) || (list.size() <= 0) ) ? 0 : list.size();
if (size == 0) {
System.out.println("Didn't find any MY_TABLE records");
} else {
System.out.println("Found [" + size + "] MY_TABLE records");
}
em.close();
emf.close();
}
}
[3]例外
javax.persistence.PersistenceException: [PersistenceUnit: MyPersistenceUnit] Unable to build EntityManagerFactory
Caused by: org.hibernate.HibernateException: Wrong column type in public.my_table for column id. Found: serial, expected: int8
更改
@GeneratedValue(strategy = GenerationType.IDENTITY, generator = "foo")
至
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "foo")
否則它將不使用順序策略,而是使用身份策略。
並嘗試刪除PK列的默認值,並將該列定義為bigint,因為int只有4個字節,而Java long有8個字節。
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