用jpa/hibernate访问辅助数据库数据的PlayFramework

Question

我正在尝试将第二个数据库连接到用PlayFramework2编写的written应用程序。

我正确地配置了我的应用程序。我已经添加了第二个源callec crm。这是我的控制台日志：

--- (RELOAD) ---

[info] play - datasource [jdbc:mysql://localhost/svp] bound to JNDI as DefaultDS
[info] play - datasource [jdbc:mysql://192.168.0.4/scrm_customer] bound to JNDI as CRM
[info] play - database [default] connected at jdbc:mysql://localhost/svp
[info] play - database [CRM] connected at jdbc:mysql://192.168.0.4/scrm_customer
[info] play - Application started (Dev)

我在我的persistence.xml中添加了以下内容：

<persistence-unit name="CRM" transaction-type="RESOURCE_LOCAL">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <non-jta-data-source>CRM</non-jta-data-source>
</persistence-unit>

我的配置是：

 db.default.jndiName=DefaultDS
 db.default.driver=com.mysql.jdbc.Driver
 db.default.url="jdbc:mysql://localhost/svp"
 db.default.user=root
 db.CRM.jndiName=CRM
 db.CRM.driver=com.mysql.jdbc.Driver
 db.CRM.url="jdbc:mysql://192.168.0.4/scrm_customer"
 db.CRM.user=root
 db.default.logStatements=true
 jpa.default=defaultPersistenceUnit

但是，当我试图使用代码从第二个db获取一些数据时，如下所示：

List<Customer> allCustomers = (List<Customer>) JPA.em("CRM")
        .createQuery("FROM Customer", Customer.class)
        .getResultList();

我收到了一个错误：

[error] play - Cannot invoke the action, eventually got an error: java.lang.RuntimeException: No JPA EntityManagerFactory configured for name [CRM]
[error] application - 

! @6kd0136e7 - Internal server error, for (GET) [/SupraADMIN/klienci] ->

play.api.Application$$anon$1: Execution exception[[RuntimeException: No JPA EntityManagerFactory configured for name [CRM]]]
        at play.api.Application$class.handleError(Application.scala:293) ~[play_2.10-2.2.4.jar:2.2.4]
        at play.api.DefaultApplication.handleError(Application.scala:399) [play_2.10-2.2.4.jar:2.2.4]
        at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$3$$anonfun$applyOrElse$3.apply(PlayDefaultUpstreamHandler.scala:264) [play_2.10-2.2.4.jar:2.2.4]
        at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$3$$anonfun$applyOrElse$3.apply(PlayDefaultUpstreamHandler.scala:264) [play_2.10-2.2.4.jar:2.2.4]
        at scala.Option.map(Option.scala:145) [scala-library.jar:na]
        at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$3.applyOrElse(PlayDefaultUpstreamHandler.scala:264) [play_2.10-2.2.4.jar:2.2.4]
Caused by: java.lang.RuntimeException: No JPA EntityManagerFactory configured for name [CRM]
        at play.db.jpa.JPA.em(JPA.java:34) ~[play-java-jpa_2.10-2.2.4.jar:2.2.4]
        at models.Customer.getCRMList(Customer.java:124) ~[na:na]
        at controllers.admin.CMS.Customers(CMS.java:157) ~[na:na]
        at admin.Routes$$anonfun$routes$1$$anonfun$applyOrElse$24$$anonfun$apply$24.apply(routes_routing.scala:429) ~[na:na]
        at admin.Routes$$anonfun$routes$1$$anonfun$applyOrElse$24$$anonfun$apply$24.apply(routes_routing.scala:429) ~[na:na]
        at play.core.Router$HandlerInvoker$$anon$7$$anon$2.invocation(Router.scala:183) ~[play_2.10-2.2.4.jar:2.2.4]
[error] application - REGUEST: GET /SupraADMIN/klienci GENERATED ERROR: @6kd0136e7: Execution exception in /home/korbeldaniel/Aplikacje/Eclipse/SVP/modules/common/app/models/Customer.java:124

我错过了什么？我查过官方文件，但没有发现有用的东西。

请帮帮忙

Answer 1

用以下注释注释控制器方法：

@Transactional(value = "CRM", readOnly = true)

在控制器方法中执行：

JPA.em().createQuery("FROM Customer", Customer.class).getResultList();

或者如果您不想使用注释：

List<Customer> customers = JPA.withTransaction("CRM", true, new Function0<List<Customer>>() {

        @Override
        public List<Customer> apply() throws Throwable {
            return JPA.em().createQuery("FROM Customer", Customer.class).getResultList();
        }
    });

我强烈建议使用JPA.withTransactionAsync。