如何在Play框架2.3x中集成JPA

Question

我正在使用play framework 2.3x 导入play.db.jpa.JPA我遇到问题;在我的Controller类Application.java中，或者不是jpa.em（）也不在控制器类

中工作

我的build.scala类是

 libraryDependencies ++= Seq(
  javaJdbc,
  javaEbean,
  cache,
  javaWs,
  "mysql" % "mysql-connector-java" % "5.1.18",
  javaJpa.exclude("org.hibernate.javax.persistence", "hibernate-jpa-2.0-api"),
  "org.hibernate" % "hibernate-entitymanager" % "4.3.9.Final",

  "com.google.code.gson" % "gson" % "2.2"
)

我的conf / META-INF / persistence.xml

<?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="defaultPersistenceUnit" transaction-type="RESOURCE_LOCAL">
        <provider>org.hibernate.ejb.HibernatePersistence</provider>
        <non-jta-data-source>DefaultDS</non-jta-data-source>
        <properties>
            <property name="hibernate.dialect" value="org.hibernate.dialect.MySQLInnoDBDialect"/>
        </properties>
    </persistence-unit>

</persistence>

我的模型类是

@javax.persistence.Entity
@Table(name = "customer")
public class Customer extends Model {
    public static Finder<Integer, Customer> find = new Finder<Integer, Customer>(
            Integer.class, Customer.class
    );

    @Id
    @Column(name = "cid")
    int cid;
    @Column(name = "name")
    public String name;

    public int getCid() {
        return cid;
    }

    public void setCid(int cid) {
        this.cid = cid;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public static List<Customer> findAll() {
        return find.all();
    }
}