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我想在同一个数据库但不同的模式中有一个单独的持久性上下文。我使用 Eclipselink 作为 JPA 实现，数据库是 PostgreSQL。

数据库应包含 2 个模式：

--txDBS
    \_data.scheme
    \_security.scheme

请您指导我如何在persistence.xml 中的一个DBS 中为每个模式声明一个持久性单元？

我现在有：

持久性.xml：

<?xml version="1.0" encoding="UTF-8" ?>
<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" xmlns="http://java.sun.com/xml/ns/persistence">

    <persistence-unit name="txPersistUnit" transaction-type="RESOURCE_LOCAL">
        <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>

        <exclude-unlisted-classes>false</exclude-unlisted-classes>
        <properties>
            <property name="javax.persistence.target-database" value="PostgreSQL"/>
            <property name="eclipselink.cache.shared.default" value="true"/>
            <property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver"/>
            <property name="javax.persistence.jdbc.url"
                      value="jdbc:postgresql://localhost:5432/TxSolution?charSet=LATIN2"/>
            <property name="javax.persistence.jdbc.user" value="txman"/>
            <property name="javax.persistence.jdbc.password" value="txman"/>

            <!-- EclipseLink should create the database schema automatically -->
            <property name="eclipselink.ddl-generation" value="create-or-extend-tables"/>
            <property name="eclipselink.ddl-generation.output-mode"
                      value="database"/>
        </properties>
    </persistence-unit>
</persistence>

所以我想做：

@Produces
​@PicketLink
​@PersistenceContext
​private EntityManager dataEm(unitName = "data.scheme.unit");

@Produces
​@PicketLink
​@PersistenceContext
​private EntityManager secEm(unitName = "security.scheme.unit");
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