【发布时间】:2016-01-27 12:00:11
【问题描述】:
我正在尝试在 Microsoft SQL Server 2012 的数据库中从 Java 自动创建表。我正在使用 JPA,并且有我的 persistence.xml:
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" 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">
<persistence-unit name="PU" transaction-type="JTA">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<jta-data-source>DS</jta-data-source>
<class>msg.Message</class>
<class>msg.Response</class>
<properties>
<property name="hibernate.connection.driver_class" value="com.microsoft.jdbc.sqlserver.SQLServerDriver"/>
<property name="hibernate.dialect" value="org.hibernate.dialect.SQLServerDialect"/>
<property name="hibernate.show_sql" value="true"/>
<property name="hibernate.hbm2ddl.auto" value="create"/>
</properties>
</persistence-unit>
glassfish ping 成功的 JDBC 连接池(附加属性
网址:jdbc:sqlserver://127.0.0.1;databaseName=test;instanceName=SQLEXPRESS; 用户:用户 密码:密码 端口号:1433)
Message.java 的头文件
@Entity
@Table(name = "message", schema = "xxx_msg")
public class Message implements Serializable {
@Id
@Basic(optional = false)
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "id")
private Integer id;
@JoinColumn(name = "sender_id", referencedColumnName = "id")
@ManyToOne
private Person senderId;
构建或部署没有错误,数据库已启动但未创建表
【问题讨论】:
标签: java sql-server hibernate jpa glassfish