如何在Play框架中使用Jpa？

问题描述 投票：1回答：1

我正在使用Play框架和JPA来存储数据库，但是有些问题即将来临：

我在做什么是我将用户存储到数据库但异常即将到来。

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");
        }

    }

}

型号是：

@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);

    }

和例外是：

[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]

任何人都可以给我一个想法来解决这个问题，是否有必要在运行应用程序之前创建sql文件

提出一些想法！

persistence.xml中

<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：

application.conf

它正在创建数据库，但它不是在项目中创建进化文件。

java hibernate jpa playframework playframework-2.1
1个回答
0
投票

要启用架构自动创建/自动更新，您需要编辑持久性单元配置文件（通常为conf/META-INF/persistence.xml）。

您需要将所需的值设置为属性hibernate.hbm2ddl.auto（查看Hibernate文档中的可用值）。您可以从值update开始，这将使Hibernate创建架构（如果它不存在）并在它与您的实体不匹配时更新它。

例如：

<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.hbm2ddl.auto" value="update" />
    </properties>
</persistence-unit>
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