我正在使用带有 JPA 的 Play 框架来存储数据库，但出现了一些问题：
我在做什么是将用户存储到数据库中，但异常即将到来。
我的控制器是：
public class Application extends Controller {
@Transactional(readOnly=true)
public static Result index() {
Form<Userd> ud=form(Userd.class);
return ok(index.render(ud));
}
@Transactional
public static Result enter(){
Form<Userd> crForm = form(Userd.class).bindFromRequest();
if(crForm.hasErrors()){
return badRequest(index.render(crForm));
}
else{
crForm.get().save();
return ok("value saved");
}
}
}
型号为：
@Entity
public class Userd {
@Required
public String name;
@Email
public String email;
@Id
public Long empid;
@Required
public String password;
public static Userd findById(Long id) {
return JPA.em().find(Userd.class,id );
}
public Userd() {
}
public Userd(String name, String email, Long empid, String password) {
this.name = name;
this.email = email;
this.empid = empid;
this.password = password;
}
public void save(){
JPA.em().persist(this);
}
例外是：
[RollbackException: Error while committing the transaction]
终端屏幕是：
[info] play - datasource [jdbc:h2:mem:play] bound to JNDI as DefaultDS
[info] play - database [default] connected at jdbc:h2:mem:play
[info] play - Application started (Dev)
[error] o.h.u.JDBCExceptionReporter - Table "USERD" not found; SQL statement:
insert into Userd (email, name, password, empid) values (?, ?, ?, ?) [42102-168]
[error] application -
! @6fiph0k4b - Internal server error, for (POST) [/login] ->
play.api.Application$$anon$1: Execution exception[[RollbackException: Error while committing the transaction]]
at play.api.Application$class.handleError(Application.scala:289) ~[play_2.10.jar:2.1.2]
at play.api.DefaultApplication.handleError(Application.scala:383) ~[play_2.10.jar:2.1.2]
at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$play$core$server$netty$PlayDefaultUpstreamHandler$$handle$1$1.apply(PlayDefaultUpstreamHandler.scala:143) ~[play_2.10.jar:2.1.2]
at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$play$core$server$netty$PlayDefaultUpstreamHandler$$handle$1$1.apply(PlayDefaultUpstreamHandler.scala:139) ~[play_2.10.jar:2.1.2]
at play.api.libs.concurrent.PlayPromise$$anonfun$extend1$1.apply(Promise.scala:113) ~[play_2.10.jar:2.1.2]
at play.api.libs.concurrent.PlayPromise$$anonfun$extend1$1.apply(Promise.scala:113) ~[play_2.10.jar:2.1.2]
javax.persistence.RollbackException: Error while committing the transaction
at org.hibernate.ejb.TransactionImpl.commit(TransactionImpl.java:93) ~[hibernate-entitymanager-3.6.9.Final.jar:3.6.9.Final]
at play.db.jpa.JPA.withTransaction(JPA.java:107) ~[play-java-jpa_2.10.jar:2.1.2]
at play.db.jpa.TransactionalAction.call(TransactionalAction.java:14) ~[play-java-jpa_2.10.jar:2.1.2]
at play.core.j.JavaAction$$anon$2.apply(JavaAction.scala:80) ~[play_2.10.jar:2.1.2]
at play.core.j.JavaAction$$anon$2.apply(JavaAction.scala:79) ~[play_2.10.jar:2.1.2]
at play.libs.F$Promise$PromiseActor.onReceive(F.java:425) ~[play_2.10.jar:2.1.2]
Caused by: javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not insert: [models.Userd]
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1387) ~[hibernate-entitymanager-3.6.9.Final.jar:3.6.9.Final]
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1315) ~[hibernate-entitymanager-3.6.9.Final.jar:3.6.9.Final]
at org.hibernate.ejb.TransactionImpl.commit(TransactionImpl.java:81) ~[hibernate-entitymanager-3.6.9.Final.jar:3.6.9.Final]
at play.db.jpa.JPA.withTransaction(JPA.java:107) ~[play-java-jpa_2.10.jar:2.1.2]
at play.db.jpa.TransactionalAction.call(TransactionalAction.java:14) ~[play-java-jpa_2.10.jar:2.1.2]
at play.core.j.JavaAction$$anon$2.apply(JavaAction.scala:80) ~[play_2.10.jar:2.1.2]
Caused by: org.hibernate.exception.SQLGrammarException: could not insert: [models.Userd]
at org.hibernate.exception.SQLStateConverter.convert(SQLStateConverter.java:92) ~[hibernate-core-3.6.9.Final.jar:3.6.9.Final]
at org.hibernate.exception.JDBCExceptionHelper.convert(JDBCExceptionHelper.java:66) ~[hibernate-core-3.6.9.Final.jar:3.6.9.Final]
at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:2454) ~[hibernate-core-3.6.9.Final.jar:3.6.9.Final]
at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:2874) ~[hibernate-core-3.6.9.Final.jar:3.6.9.Final]
at org.hibernate.action.EntityInsertAction.execute(EntityInsertAction.java:79) ~[hibernate-core-3.6.9.Final.jar:3.6.9.Final]
at org.hibernate.engine.ActionQueue.execute(ActionQueue.java:273) ~[hibernate-core-3.6.9.Final.jar:3.6.9.Final]
Caused by: org.h2.jdbc.JdbcSQLException: Table "USERD" not found; SQL statement:
insert into Userd (email, name, password, empid) values (?, ?, ?, ?) [42102-168]
at org.h2.message.DbException.getJdbcSQLException(DbException.java:329) ~[h2.jar:1.3.168]
at org.h2.message.DbException.get(DbException.java:169) ~[h2.jar:1.3.168]
at org.h2.message.DbException.get(DbException.java:146) ~[h2.jar:1.3.168]
at org.h2.command.Parser.readTableOrView(Parser.java:4770) ~[h2.jar:1.3.168]
at org.h2.command.Parser.readTableOrView(Parser.java:4748) ~[h2.jar:1.3.168]
at org.h2.command.Parser.parseInsert(Parser.java:958) ~[h2.jar:1.3.168]
任何人都可以给我解决这个问题的想法吗？是否有必要在运行应用程序之前创建 sql 文件
给点主意！
持久性.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"/>
<property name="hibernate.hbm2ddl.auto" value="update"/>
</properties>
</persistence-unit>
</persistence>
application.conf：你可以在这个链接上看到：
它正在创建数据库，但没有在项目中创建进化文件。