playframework JPA.em（）。merge（this） [英] playframework JPA.em().merge(this)
本文介绍了playframework JPA.em（）。merge（this）的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我试图将我的表单插入到h2数据库中。
我使用hibernate和jpa。这是代码
包模型;
import java.util。*;
import javax.persistence。*;
import play.data.format。*;
import play.data.validation。*;
import play.db.jpa。*;
@Entity
公共类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;
/ **
*插入新的事件提交。
* /
@Transactional
public void toDataBase（）{
// persist object - add to entity manager
JPA.em（）。merge（this）;
JPA.em（）。persist（this）;
}
}
我正在面对一个问题。这是一个错误：
$ p $ [PersistenceException：org.hibernate.exception.SQLGrammarException：无法准备语句]
它看起来像使用ebean而不是jpa / hibernate。我从堆栈跟踪访问：
[info] play - datasource [jdbc：h2：mem：SupraCIRS]绑定到JNDI为DefaultDS
[info] play - database [default]连接在jdbc：h2：mem：SupraCIRS
[info] play - 应用程序已启动（Dev）
[错误] ohejsSqlExceptionHelper - KolumnaMEDICALINC0_ ._EBEAN_INTERCEPTnie istnieje
未找到MEDICALINC0 _._ EBEAN_INTERCEPT列; SQL语句：
选择medicalinc0_.id作为id1_0_0_，medicalinc0 _._ ebean_intercept作为_ebean_i2_0_0_，medicalinc0_.age作为age3_0_0_，medicalinc0_.incident_type作为incident4_0_0_，medicalinc0_.month作为month5_0_0_，medicalinc0_.place作为place6_0_0_，medicalinc0_.unit作为unit7_0_0_来自MedicalIncident medicalinc0_ where medicalinc0_.id =？ [42122-172]
[error] play - 无法调用动作，最终出错：javax.persistence.PersistenceException：org.hibernate.exception.SQLGrammarException：无法准备语句
[error]应用程序 -
！ @ 6ghh1j2nn - 内部服务器错误，用于（POST）[/ incydent_medyczny / zapisz] - >
play.api.Application $$ anon $ 1：执行异常[[PersistenceException：org.hibernate.exception.SQLGrammarException：无法准备语句]]
at play.api.Application $ class .handleError（Application.scala：293）〜[play_2.10.jar：2.2.1]
at play.api.DefaultApplication.handleError（Application.scala：399）[play_2.10.jar：2.2.1 ]
at play.core.server.netty.PlayDefaultUpstreamHandler $$ anonfun $ 2 $$ anonfun $ applyOrElse $ 3.apply（PlayDefaultUpstreamHandler.scala：261）[play_2.10.jar：2.2.1]
at play.core.server.netty.PlayDefaultUpstreamHandler $$ anonfun $ 2 $$ anonfun $ applyOrElse $ 3.apply（PlayDefaultUpstreamHandler.scala：261）[play_2.10.jar：2.2.1]
at scala.Option.map（在play.core.server.netty.PlayDefaultUpstreamHandler $$ anonfun $ 2.applyOrElse（PlayDefaultUpstreamHandler.scala：261）[play_2.10.jar：2.2] [scala-library.jar：na]
.1]
导致：javax.persistence.PersistenceException：org.hibernate.exception.SQLG rammarException：无法在org.hibernate.ejb.AbstractEntityManagerImpl.convert（AbstractEntityManagerImpl.java:1387）〜[hibernate-entitymanager-4.2.6.Final.jar：4.2.6.Final]中准备语句
在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]
导致：org.hibernate.exception.SQLGrammarException：无法在org.hibernate.exception.internal.SQLStateC中准备语句
onversionDelegate.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]
在org.hibernate.loader.Loader.prepareQueryStatement（Loader.java:1859）〜[hibernate-core-4.2.6.Final.jar：4.2.6.Final ]
导致：org.h2.jdbc.JdbcSQLException：KolumnaMEDICALINC0 _._ EB EAN_INTERCEPTnie istnieje
未找到MEDICALINC0 _._ EBEAN_INTERCEPT列; SQL语句：
选择medicalinc0_.id作为id1_0_0_，medicalinc0 _._ ebean_intercept作为_ebean_i2_0_0_，medicalinc0_.age作为age3_0_0_，medicalinc0_.incident_type作为incident4_0_0_，medicalinc0_.month作为month5_0_0_，medicalinc0_.place作为place6_0_0_，medicalinc0_.unit作为unit7_0_0_来自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conf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 = Defa ultDS
jpa.default = defaultPersistenceUnit
ebeanEnabled = false
logger.root =错误
logger.play = INFO
logger.application = DEBUG
http.port = 8081
文件build.sbt
import play.Project._
name：=SupraCIRS
版本：=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.or g / 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 =defaultPersistenceUnittransaction-type =RESOURCE_LOCAL>
< provider> org.hibernate.ejb.HibernatePersistence< / provider>
<非-jta-data-source> DefaultDS< /非-jta-data-source>
<属性>
< property name =hibernate.dialectvalue =org.hibernate.dialect.H2Dialect/>
< / properties>
< / persistence-unit>
< /持久性>
请帮忙
解决方案
问题在这里： libraryDependencies ++ = Seq（javaJdbc，javaEbean）。删除 javaEbean 部分！这也应该解决您的其他问题。
I am trying to Insert my form to h2 database. I am using hibernate and jpa. Here is the code
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);
}
}
I am faceing an issue. Here is an error:
[PersistenceException: org.hibernate.exception.SQLGrammarException: could not prepare statement]
Its look like it using ebean instead of jpa/hibernate. I am guesting from stack trace:
[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]
I dont understand: I've precise jpa in conf:
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
file 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>
Plese help
解决方案
The problem is here: libraryDependencies ++= Seq(javaJdbc, javaEbean). Remove the javaEbean part! This should also solve all your other problems.
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