Hibernate playframework JPA.em（）.merge（此）

我正在尝试将我的表单插入h2数据库。 我正在使用hibernate和jpa。这是密码

package models;
import java.util.*;
import javax.persistence.*;
import play.data.format.*;
import play.data.validation.*;
import play.db.jpa.*;

@Entity
public class MedicalIncident   {

    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    public int id;

    @Constraints.Required
    public String month;

    @Constraints.Required
    public String place;

    @Constraints.Required
    public String unit;

    @Constraints.Required
    public String incident_type;

    @Constraints.Required
    public int age;


    /**
     * Insert this new incident submission.
     */
    @Transactional
    public void toDataBase() {
        // persist object - add to entity manager
        JPA.em().merge(this);
        JPA.em().persist(this);
    }


}

我面临一个问题。这里有一个错误：

 [PersistenceException: org.hibernate.exception.SQLGrammarException: could not prepare statement] 

它看起来像是使用ebean而不是jpa/hibernate。我是stack trace的访客：

[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] o.h.e.j.s.SqlExceptionHelper - Kolumna "MEDICALINC0_._EBEAN_INTERCEPT" nie istnieje
Column "MEDICALINC0_._EBEAN_INTERCEPT" not found; SQL statement:
select medicalinc0_.id as id1_0_0_, medicalinc0_._ebean_intercept as _ebean_i2_0_0_, medicalinc0_.age as age3_0_0_, medicalinc0_.incident_type as incident4_0_0_, medicalinc0_.month as month5_0_0_, medicalinc0_.place as place6_0_0_, medicalinc0_.unit as unit7_0_0_ from MedicalIncident medicalinc0_ where medicalinc0_.id=? [42122-172]
[error] play - Cannot invoke the action, eventually got an error: javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not prepare statement
[error] application - 

! @6ghh1j2nn - Internal server error, for (POST) [/incydent_medyczny/zapisz] ->

play.api.Application$$anon$1: Execution exception[[PersistenceException: org.hibernate.exception.SQLGrammarException: could not prepare statement]]
        at play.api.Application$class.handleError(Application.scala:293) ~[play_2.10.jar:2.2.1]
        at play.api.DefaultApplication.handleError(Application.scala:399) [play_2.10.jar:2.2.1]
        at play.core.server.netty.PlayDefaultUpstreamHandler$$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$2.applyOrElse(PlayDefaultUpstreamHandler.scala:261) [play_2.10.jar:2.2.1]
Caused by: javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not prepare statement
        at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1387) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1310) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1316) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.ejb.AbstractEntityManagerImpl.merge(AbstractEntityManagerImpl.java:898) ~[hibernate-entitymanager-4.2.6.Final.jar:4.2.6.Final]
        at models.MedicalIncident.toDataBase(MedicalIncident.java:40) ~[na:na]
        at controllers.MedicalIncidents.submit(MedicalIncidents.java:50) ~[na:na]
Caused by: org.hibernate.exception.SQLGrammarException: could not prepare statement
        at org.hibernate.exception.internal.SQLStateConversionDelegate.convert(SQLStateConversionDelegate.java:123) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:49) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:125) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.engine.jdbc.internal.StatementPreparerImpl$StatementPreparationTemplate.prepareStatement(StatementPreparerImpl.java:188) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.engine.jdbc.internal.StatementPreparerImpl.prepareQueryStatement(StatementPreparerImpl.java:159) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
        at org.hibernate.loader.Loader.prepareQueryStatement(Loader.java:1859) ~[hibernate-core-4.2.6.Final.jar:4.2.6.Final]
Caused by: org.h2.jdbc.JdbcSQLException: Kolumna "MEDICALINC0_._EBEAN_INTERCEPT" nie istnieje
Column "MEDICALINC0_._EBEAN_INTERCEPT" not found; SQL statement:
select medicalinc0_.id as id1_0_0_, medicalinc0_._ebean_intercept as _ebean_i2_0_0_, medicalinc0_.age as age3_0_0_, medicalinc0_.incident_type as incident4_0_0_, medicalinc0_.month as month5_0_0_, medicalinc0_.place as place6_0_0_, medicalinc0_.unit as unit7_0_0_ from MedicalIncident medicalinc0_ where medicalinc0_.id=? [42122-172]
        at org.h2.message.DbException.getJdbcSQLException(DbException.java:329) ~[h2.jar:1.3.172]
        at org.h2.message.DbException.get(DbException.java:169) ~[h2.jar:1.3.172]
        at org.h2.message.DbException.get(DbException.java:146) ~[h2.jar:1.3.172]
        at org.h2.expression.ExpressionColumn.optimize(ExpressionColumn.java:144) ~[h2.jar:1.3.172]
        at org.h2.expression.Alias.optimize(Alias.java:52) ~[h2.jar:1.3.172]
        at org.h2.command.dml.Select.prepare(Select.java:808) ~[h2.jar:1.3.172]

我不明白：我在形态中有精确的jpa：

application.name="SupraCIRS"
application.lang.cookie=SUPRACIRS_LANG
application.secret="r5Wd;ABfd7;e^F3p9WUl2132Y13v213g12K3ljoIdsg`MU8`YL:x`8KAhgao21ofwb?iQJe6hL0I6liIg121211"
application.langs="pl"

db.default.driver=org.h2.Driver
db.default.url="jdbc:h2:mem:SupraCIRS"
db.default.jndiName=DefaultDS

jpa.default=defaultPersistenceUnit

ebeanEnabled=false
logger.root=ERROR
logger.play=INFO
logger.application=DEBUG
http.port=8081

文件build.sbt

import play.Project._

name := """SupraCIRS"""

version := "1.0-Alfa"

libraryDependencies ++= Seq(
    "org.webjars" %% "webjars-play" % "2.2.0", 
    "org.webjars" % "bootstrap" % "2.3.1")

libraryDependencies ++= Seq(javaJdbc, javaEbean)

libraryDependencies ++= Seq(
  "org.hibernate" % "hibernate-entitymanager" % "4.2.6.Final",
  "mysql" % "mysql-connector-java" % "5.1.18",
  jdbc,
  javaCore,
  javaJdbc,
  javaJpa
)

playJavaSettings

conf/META-INF/persistence.xml

<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.H2Dialect"/>
        </properties>
    </persistence-unit>

</persistence>

org.hibernate.ejb.HibernatePersistence
默认值

---

请提供帮助：

libraryDependencies++=Seq（javaJdbc，javaEbean）

。删除

javaEbean

部分！这还可以解决所有其他问题。

解决方案：

我有一个干净的拉链。 并在新的基础上创建项目。

---

然后，我将旧项目中的dir app configuration public放入新创建的项目中

您是否尝试过像这里描述的那样添加对hibernate的支持：？是的，我遵循了Documentation，但我在您上面发布的conf文件中没有看到这些说明..这是什么意思？我真的把项目中的依赖关系搞砸了。我不知道为什么清理build.sbt没有启用。我一直都有同样的问题。因此，我在build.sbt中创建了一个没有ebean的新项目。我还有一个新的剧本（使用干净的本地存储库）。所有这些都帮助了我。如果您有更好的解决方案，请与我分享！