Question

我面临着将eclipselink集成到play框架中的一些问题。当我键入运行时，服务器启动，我可以连接到我的应用程序，没有任何错误。

但是：之后当我（作为一个例子）强制重新编译并稍加改动时，应用程序就会出错（在em.persist上）：

play.api.Application$$anon$1: Execution exception[[IllegalArgumentException: Object: model.Car@1a5f139c is not a known entity type.]]

我做了什么（保持简单）汽车模型：

package model;

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "car")
public class Car {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private int carId;

    private String name;

    public String getName() {
        return this.name;
    }

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

}

我的build.sbt看起来像这样：

name := "WebTest"

version := "1.0-SNAPSHOT"

libraryDependencies ++= Seq(
 "org.eclipse.persistence" % "eclipselink" % "2.5.1",
 "mysql" % "mysql-connector-java" % "5.1.18",
  javaJdbc,
  javaEbean,
  cache
)     


play.Project.playJavaSettings

application.conf（部分地）：

db.default.jndiName=DefaultDS

db.default.driver=com.mysql.jdbc.Driver
db.default.url="jdbc:mysql://localhost/playjpa?characterEncoding=UTF-8"
db.default.user=xxxx
db.default.password=xxxx

jpa.default=defaultPersistenceUnit

在conf.META-INFO中解析我的persistence.xml（之前删除了创建）：

<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">
        <class>model.Car</class>

        <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
        <non-jta-data-source>DefaultDS</non-jta-data-source>
        <exclude-unlisted-classes>false</exclude-unlisted-classes>
         <properties>
            <property name="eclipselink.ddl-generation" value="create-or-extend-tables" />
            <property name="eclipselink.ddl-generation.output-mode" value="database" />
        </properties>
    </persistence-unit>
</persistence>

现在执行调用的控制器：

public class Application extends Controller {

    private static final String PERSISTENCE_UNIT_NAME = "defaultPersistenceUnit";
    private static EntityManagerFactory factory;

    public static Result index() {
        factory = Persistence.createEntityManagerFactory(PERSISTENCE_UNIT_NAME);
        EntityManager em = factory.createEntityManager();
        em.getTransaction().begin();

        Car car = new Car();
        car.setName("Supercar");

        em.persist(car);
        em.getTransaction().commit();
        em.close();

        return ok(index.render("Your new application is ready."));
    }

}

提前致谢！

Answer 1

自己解决了这个问题：

关键是使用@Transactional进行注释。但是在build.sbt必须像这样修改之前

 "org.eclipse.persistence" % "eclipselink" % "2.5.1",
 "mysql" % "mysql-connector-java" % "5.1.18",
  javaJdbc,
  javaJpa

在游戏机中你必须“重新加载”。并且只有“eclipse”再次导入项目（至少我的日食不喜欢它）你可以导入

import play.db.jpa.Transactional;

结束

JPA.em()

你可以获得实体经理。

在更改和重新加载时使用eclipselink错误播放框架

1 个答案: