0

我是 Openjpa 的新手。我需要为名为 Draco-PU 的持久性单元显式创建一个 EntityManager 对象。我定义了以下持久性 xml。请帮忙！

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="1.0" 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_1_0.xsd">
<persistence-unit name="Draco-PU" >
<provider>org.apache.openjpa.persistence.PersistenceProviderImpl</provider>
<jta-data-source>DracoDataSource</jta-data-source>
<non-jta-data-source>DracoUnmanagedDataSource</non-jta-data-source>
<class>dk.tdc.soa.smo.draco.db.model.DslServiceEntity</class>
<class>dk.tdc.soa.smo.draco.db.model.DslServiceCatalogEntity</class>
<class>dk.tdc.soa.smo.draco.db.model.History</class>
<class>dk.tdc.soa.smo.draco.db.model.ConfigEntity</class>
<properties>
  <property name="openjpa.Log" value="DefaultLevel=WARNING, Runtime=WARNING, Tool=WARNING, SQL=WARNING" />
  <!-- <property name="openjpa.jdbc.DBDictionary" value="org.apache.openjpa.jdbc.sql.OracleDictionary" />  -->     
  <property name="openjpa.jdbc.SynchronizeMappings" value="buildSchema(ForeignKeys=true)" />
</properties>
</persistence-unit>
</persistence>
4

1 回答 1

1

两周前我是新来的，所以我的回答物有所值，但我确实喜欢这样：

EntityManager entityManager = 
Persistence.createEntityManagerFactory("nameofyourpersistenceunit").createEntityManager();
于 2012-10-22T09:45:00.653 回答