playframework org.hibernate.PersistentObjectException：传递给persist的分离实体：models

Question

我正在尝试使用JPA和H2数据库设置playframework 2.2的应用程序。 我选择了这个： http ： //www.playframework.com/documentation/2.2.x/JavaJPA和computer-database-jpa示例项目

我收到了一个错误：

 org.hibernate.PersistentObjectException: detached entity passed to persist: models.

当我运行我的项目时，我收到了这个错误：

---

堆栈跟踪

---

        --- (Running the application from SBT, auto-reloading is enabled) ---

[info] play - Listening for HTTP on /0:0:0:0:0:0:0:0:8081

(Server started, use Ctrl+D to stop and go back to the console...)

[info] play - datasource [jdbc:h2:mem:SupraCIRS] bound to JNDI as DefaultDS
[info] play - database [default] connected at jdbc:h2:mem:SupraCIRS
[info] play - Application started (Dev)
[error] play - Cannot invoke the action, eventually got an error: javax.persistence.PersistenceException: org.hibernate.PersistentObjectException: detached entity passed to persist: models.MedicalIncident
[error] application - 

! @6gh420nj2 - Internal server error, for (POST) [/incydent_medyczny/zapisz] ->

play.api.Application$$anon$1: Execution exception[[PersistenceException: org.hibernate.PersistentObjectException: detached entity passed to persist: models.MedicalIncident]]
        at play.api.Application$class.handleError(Application.scala:293) ~[play_2.10.jar:2.2.1]
        at play.api.DefaultApplication.handleError(Application.scala:399) [play_2.10.jar:2.2.1]
        at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$2$$anonfun$applyOrElse$3.apply(PlayDefaultUpstreamHandler.scala:261) [play_2.10.jar:2.2.1]
        at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$2$$anonfun$applyOrElse$3.apply(PlayDefaultUpstreamHandler.scala:261) [play_2.10.jar:2.2.1]
        at scala.Option.map(Option.scala:145) [scala-library.jar:na]
        at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$2.applyOrElse(PlayDefaultUpstreamHandler.scala:261) [play_2.10.jar:2.2.1]
Caused by: javax.persistence.PersistenceException: org.hibernate.PersistentObjectException: detached entity passed to persist: models.MedicalIncident
        at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1387) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1310) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1316) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.ejb.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:881) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at models.MedicalIncident.toDataBase(MedicalIncident.java:40) ~[na:na]
        at controllers.MedicalIncidents.submit(MedicalIncidents.java:41) ~[na:na]
Caused by: org.hibernate.PersistentObjectException: detached entity passed to persist: models.MedicalIncident
        at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:141) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:78) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.internal.SessionImpl.firePersist(SessionImpl.java:853) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:827) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:831) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.ejb.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:875) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
-

-- (Running the application from SBT, auto-reloading is enabled) ---

[info] play - Listening for HTTP on /0:0:0:0:0:0:0:0:8081

(Server started, use Ctrl+D to stop and go back to the console...)

[info] Compiling 1 Java source to /home/daniel/play-2.2.1/SupraCIRS/target/scala-2.10/classes...

[info] Compiling 3 Scala sources and 1 Java source to /home/daniel/play-2.2.1/SupraCIRS/target/scala-2.10/classes...
[info] play - datasource [jdbc:h2:mem:play] bound to JNDI as DefaultDS
[info] play - database [default] connected at jdbc:h2:mem:play
[error] application - 

! @6gh36abca - Internal server error, for (GET) [/] ->

play.api.UnexpectedException: Unexpected exception[PersistenceException: [PersistenceUnit: defaultPersistenceUnit] Unable to build EntityManagerFactory]
        at play.core.ReloadableApplication$$anonfun$get$1$$anonfun$apply$1$$anonfun$1.apply(ApplicationProvider.scala:148) ~[play_2.10.jar:2.2.1]
        at play.core.ReloadableApplication$$anonfun$get$1$$anonfun$apply$1$$anonfun$1.apply(ApplicationProvider.scala:112) ~[play_2.10.jar:2.2.1]
        at scala.Option.map(Option.scala:145) ~[scala-library.jar:na]
        at play.core.ReloadableApplication$$anonfun$get$1$$anonfun$apply$1.apply(ApplicationProvider.scala:112) ~[play_2.10.jar:2.2.1]
        at play.core.ReloadableApplication$$anonfun$get$1$$anonfun$apply$1.apply(ApplicationProvider.scala:110) ~[play_2.10.jar:2.2.1]
        at scala.util.Success.flatMap(Try.scala:200) ~[scala-library.jar:na]
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: defaultPersistenceUnit] Unable to build EntityManagerFactory
        at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:924) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:899) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.ejb.HibernatePersistence.createEntityManagerFactory(HibernatePersistence.java:59) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:63) ~[hibernate-jpa-2.0-api.jar:1.0.1.Final]
        at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:47) ~[hibernate-jpa-2.0-api.jar:1.0.1.Final]
        at play.db.jpa.JPAPlugin.onStart(JPAPlugin.java:35) ~[play-java-jpa_2.10.jar:2.2.1]
Caused by: org.hibernate.AnnotationException: No identifier specified for entity: models.MedicalIncident
        at org.hibernate.cfg.InheritanceState.determineDefaultAccessType(InheritanceState.java:277) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.cfg.InheritanceState.getElementsToProcess(InheritanceState.java:224) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.cfg.AnnotationBinder.bindClass(AnnotationBinder.java:665) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.cfg.Configuration$MetadataSourceQueue.processAnnotatedClassesQueue(Configuration.java:3533) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.cfg.Configuration$MetadataSourceQueue.processMetadata(Configuration.java:3487) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.cfg.Configuration.secondPassCompile(Configuration.java:1376) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]


    //end

这是在控制器中处理表单提交的类

    /**
 * Handle the form submission.
 */
@Transactional
public static Result submit() {
    Form<MedicalIncident> filledForm = MedicalIncidentForm.bindFromRequest();

    if(filledForm.hasErrors()) {
        return badRequest(form.render(filledForm));
    } else {
        MedicalIncident created = filledForm.get();
        //return ok(confirmation.render("Twój incydent został zgłoszony poprawnie."));
        created.save();
        return ok(summary.render(created));
    }
}

这是我的模特课

    package models;

import java.util.*;
import javax.persistence.*;
import javax.persistence.Entity;
import javax.validation.*;
import play.data.validation.Constraints.*;
import play.api.db.*;
import play.db.jpa.*;
import javax.persistence.EntityManager;

@Entity
public class MedicalIncident   {

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    public Long id;

    @Required
    public String month;

    @Required
    public String place;

    @Required
    public String unit;

    @Required
    public String incident_type;

    @Required
    public int age;



    /**
     * Insert this new incident submission.
     */
    public void toDataBase() {
        JPA.em().persist(this);
    }



}

请帮助我

Answer 1

从堆栈跟踪中我假设您没有在MedicalIncident类的任何字段上放置@Id注释，但由于您尚未发布它我无法确定。

Hibernate需要@Id字段来唯一标识记录。

请参阅hibernate docs 。