java 如何将实体绑定到特定的持久性单元

How to bind an entity to a specific persistence-unit

提问by Javatar

In a web application using struts2 ejb hibernate, is it possible to tell the application to find or create an entity for a specific persistence-unit name, which is written in persistence.xmlfile, in the deployment time?

在使用 struts2 ejb hibernate 的 web 应用程序中，是否可以persistence.xml在部署时告诉应用程序为特定的持久性单元名称查找或创建实体，该名称写入文件？

I have two persistence-unit in persistence.xml, and one datasource (including two "local-tx-datasource") xml file under the jboss node.

我persistence.xml在 jboss 节点下有两个持久性单元和一个数据源（包括两个“local-tx-datasource”）xml 文件。

To clearify, I mean, I tried this;

为了澄清，我的意思是，我试过这个；

@Entity  
@PersistenceContext(unitName="MY JNDI NAME specified in persistence.xml") 
public abstract class Vehicle {

and doesnt work.. Then tried this and etc..

并不起作用..然后尝试了这个等等..

@PersistenceContext(name="MY PERSISTENCE UNIT NAME specified in persistence.xml")

@PersistenceUnit(name="MY PERSISTENCE UNIT NAME specified in persistence.xml")

and also I tried these above with the "UnitName=.." instead of "name=.." but anything is worked for me...

而且我在上面用“UnitName = ..”而不是“name = ..”尝试了这些，但任何东西都对我有用......

[SOLVED]

[解决了]

<.exclude-unlisted-classes>true<./exclude-unlisted-classes> has solved my problem

<.exclude-unlisted-classes>true<./exclude-unlisted-classes> 解决了我的问题

回答by Pascal Thivent

Update:Based on your comment (this is not what I understood from the original question), I don't think you have any other option than disabling "discovery" and listing explicitly your entities in their respective persistence unit:

更新：根据您的评论（这不是我从原始问题中所理解的），我认为除了禁用“发现”并在各自的持久性单元中明确列出您的实体之外，您别无选择：

<?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="MyPu1" transaction-type="RESOURCE_LOCAL">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <class>com.mycompany.Foo</class>
    ...
    <exclude-unlisted-classes>true</exclude-unlisted-classes>
    <properties>
      <!-- H2 in memory -->
      <property name="javax.persistence.jdbc.driver" value="org.h2.Driver"/>
      <property name="javax.persistence.jdbc.url" value="jdbc:h2:mem:test"/>
      <property name="javax.persistence.jdbc.username" value="APP"/>
      <property name="javax.persistence.jdbc.password" value="APP"/>
      <property name="hibernate.hbm2ddl.auto" value="update"/>
    </properties>
  </persistence-unit>

  <persistence-unit name="MyPu2" transaction-type="RESOURCE_LOCAL">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <class>com.mycompany.Bar</class>
    ...
    <exclude-unlisted-classes>true</exclude-unlisted-classes>
    <properties>
      <!-- Derby server -->
      <property name="javax.persistence.jdbc.driver" value="org.apache.derby.jdbc.ClientDriver"/>
      <property name="javax.persistence.jdbc.user" value="APP"/>
      <property name="javax.persistence.jdbc.password" value="APP"/>
      <property name="javax.persistence.jdbc.url" value="jdbc:derby://localhost:1527/Pu2;create=true"/>
      <property name="hibernate.hbm2ddl.auto" value="update"/>
    </properties>
  </persistence-unit>
</persistence>

I'm not aware of any syntax at the entity level allowing to assign it to a persistence unit.

我不知道实体级别的任何语法允许将其分配给持久性单元。

---

---

I'm not sure I understood what you're trying to do but if you want to get an Entity Manager for a specific persistence unit injected, you should do:

我不确定我是否理解您要执行的操作，但是如果您想为特定持久性单元注入实体管理器，您应该执行以下操作：

@Stateless
public class FooBean implements Foo {
    @PersistenceContext(unitName="MyPu1")
    EntityManager em1;

    // ...
}

@Stateless
public class FooBean implements Foo {
    @PersistenceContext(unitName="MyPu1")
    EntityManager em1;

    // ...
}

If this is not what you want, please clarify the question.

如果这不是您想要的，请澄清问题。

回答by Sylwester Gebczyk

What you're looking for is probably <exclude-unlisted-classes>true</exclude-unlisted-classes>.

您正在寻找的可能是<exclude-unlisted-classes>true</exclude-unlisted-classes>.

Check the documentation on jboss.org:

查看 jboss.org 上的文档：

In my configuration I had two databases (let's say A and B) and I wanted two separate persistence units, where one contains all entities, but one, while the other persistence unit contains the remaining entity. My persistence.xml looks like this:

在我的配置中，我有两个数据库（比方说 A 和 B），我想要两个单独的持久性单元，其中一个包含所有实体，但一个，而另一个持久性单元包含剩余的实体。我的persistence.xml 看起来像这样：

<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="A" transaction-type="RESOURCE_LOCAL">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <properties>
        <property name="hibernate.dialect" value="${hibernate.dialect}" />
    </properties>
</persistence-unit>

<persistence-unit  name="B" transaction-type="RESOURCE_LOCAL">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <exclude-unlisted-classes>true</exclude-unlisted-classes>
    <properties>
        <property name="hibernate.dialect" value="${rud.hibernate.dialect}"/>
    </properties>
    <class>com.sgk.Person</class>
</persistence-unit>

回答by Nayan Wadekar

@Pascal Thivent

@Pascal Thivent

I haven't tried using multiple EntityManager at once, but looking at above mentioned problem this may help if it works.

我还没有尝试一次使用多个 EntityManager，但是查看上面提到的问题，如果它有效，这可能会有所帮助。

@PersistenceContext(unitName="MyPu1") EntityManager em1;

@PersistenceContext(unitName="MyPu1") EntityManager em1;

@PersistenceContext(unitName="MyPu2") EntityManager em2;

@PersistenceContext(unitName="MyPu2") EntityManager em2;