使用OpenJPA进行交易无效

时间:2015-12-08 12:14:00

标签: java postgresql servlets jpa openjpa

我的servlet有一个令人沮丧的问题。这是非常基本的,我应该在我的数据库表中添加一个博客条目,其中包含3列(id,tittel(title),main(mainbody))。我可以从我的数据库加载并显示在我的jsp文件中。但是我无法将数据保存到数据库中,我无法弄清楚原因。

我使用postgres作为数据库。

这是我的“saveBlog” - 方法:

public void lagreBlogg(Blogg b){
           EntityManager em =      Persistence.createEntityManagerFactory("kimsBlogg").createEntityManager();
            EntityTransaction transaction = em.getTransaction();

           try {
            transaction.begin();
            em.persist(b);
            transaction.commit();
        } catch (Exception e) {
            transaction.rollback();

        } finally {
            em.close();
        }

    }

}

这是我的实体类:

 @Entity
 @Table(schema = "blogg", name = "blogg")
 public class Blogg {
@Id
@GeneratedValue
private int id;
private String tittel;
private String main;


public Blogg() {

}

public Blogg(String tittel, String main){
    this.tittel= tittel;
    this.main = main;
}

public String getTittel() {
    return tittel;
}

public void setTittel(String tittel) {
    this.tittel = tittel;
}

public String getMain() {
    return main;
}

public void setMain(String main) {
    this.main = main;
}
}

我的persistence.xml

<?xml version="1.0" encoding="UTF-8" ?>
<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" xmlns="http://java.sun.com/xml/ns/persistence">

<persistence-unit name="kimsBlogg">
    <provider>org.apache.openjpa.persistence.PersistenceProviderImpl
    </provider>

    <class>no.kim.model.Blogg</class>




    <properties>
        <property name="javax.persistence.jdbc.logging.level" value="INFO" />
        <property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver" />
        <property name="javax.persistence.jdbc.url" value="jdbc:postgresql://localhost:5432/postgres" />
        <property name="javax.persistence.jdbc.user" value="postgres" />
        <property name="javax.persistence.jdbc.password" value="krheggen" />
        <property name="openjpa.RuntimeUnenhancedClasses" value="supported" />
    </properties>
</persistence-unit>

这是我得到的错误:

HTTP状态500 - 只能在事务处于活动状态时执行操作。

输入例外报告

message只能在事务处于活动状态时执行操作。

说明服务器遇到内部错误,导致无法完成此请求。

例外

org.apache.openjpa.persistence.InvalidStateException:只能在事务处于活动状态时执行操作。     org.apache.openjpa.kernel.BrokerImpl.assertTransactionOperation(BrokerImpl.java:4732)     org.apache.openjpa.kernel.BrokerImpl.rollback(BrokerImpl.java:1543)     org.apache.openjpa.kernel.DelegatingBroker.rollback(DelegatingBroker.java:940)     org.apache.openjpa.persistence.EntityManagerImpl.rollback(EntityManagerImpl.java:598)     no.kim.helpers.BloggDAO.lagreBlogg(BloggDAO.java:37)     no.kim.Controller.BloggServlet.doPost(BloggServlet.java:33)     javax.servlet.http.HttpServlet.service(HttpServlet.java:650)     javax.servlet.http.HttpServlet.service(HttpServlet.java:731)     org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)

栈跟踪

引起:org.apache.openjpa.persistence.PersistenceException:错误:关系“blogg.openjpa_sequence_table”不存在   位置:28 {prepstmnt 166406485 SELECT SEQUENCE_VALUE from blogg.OPENJPA_SEQUENCE_TABLE WHERE ID =? FOR UPDATE} [code = 0,state = 42P01]     在org.apache.openjpa.jdbc.sql.DBDictionary.narrow(DBDictionary.java:5001)     at org.apache.openjpa.jdbc.sql.DBDictionary.newStoreException(DBDictionary.java:4961)     在org.apache.openjpa.jdbc.sql.SQLExceptions.getStore(SQLExceptions.java:133)     在org.apache.openjpa.jdbc.sql.SQLExceptions.getStore(SQLExceptions.java:107)     在org.apache.openjpa.jdbc.sql.SQLExceptions.getStore(SQLExceptions.java:59)     在org.apache.openjpa.jdbc.kernel.AbstractJDBCSeq.next(AbstractJDBCSeq.java:63)     在org.apache.openjpa.util.ImplHelper.generateValue(ImplHelper.java:159)     在org.apache.openjpa.util.ImplHelper.generateFieldValue(ImplHelper.java:143)     在org.apache.openjpa.jdbc.kernel.JDBCStoreManager.assignField(JDBCStoreManager.java:774)     在org.apache.openjpa.util.ApplicationIds.assign(ApplicationIds.java:492)     在org.apache.openjpa.util.ApplicationIds.assign(ApplicationIds.java:468)     at org.apache.openjpa.jdbc.kernel.JDBCStoreManager.assignObjectId(JDBCStoreManager.java:758)     在org.apache.openjpa.kernel.DelegatingStoreManager.assignObjectId(DelegatingStoreManager.java:135)     at org.apache.openjpa.kernel.StateManagerImpl.assignObjectId(StateManagerImpl.java:595)     at org.apache.openjpa.kernel.StateManagerImpl.preFlush(StateManagerImpl.java:3051)     在org.apache.openjpa.kernel.PNewState.beforeFlush(PNewState.java:44)     在org.apache.openjpa.kernel.StateManagerImpl.beforeFlush(StateManagerImpl.java:1072)     在org.apache.openjpa.kernel.BrokerImpl.flush(BrokerImpl.java:2143)     在org.apache.openjpa.kernel.BrokerImpl.flushSafe(BrokerImpl.java:2103)     在org.apache.openjpa.kernel.BrokerImpl.beforeCompletion(BrokerImpl.java:2021)     在org.apache.openjpa.kernel.LocalManagedRuntime.commit(LocalManagedRuntime.java:81)     在org.apache.openjpa.kernel.BrokerImpl.commit(BrokerImpl.java:1526)     在org.apache.openjpa.kernel.DelegatingBroker.commit(DelegatingBroker.java:932)     在org.apache.openjpa.persistence.EntityManagerImpl.commit(EntityManagerImpl.java:569)

2 个答案:

答案 0 :(得分:1)

问题在于catch方法中的" transaction.rollback();"。如果提交因异常而失败,则连接不再处于活动状态。

答案 1 :(得分:1)

  

relationship&#34; blogg.openjpa_sequence_table&#34;不存在

这似乎很简单,只需为表blogg创建一个序列,更多信息here