没有活动交易，获取无效 [英] Get is not valid without active transaction

问题描述

这是配置：

 < context：annotation-config /> 
< context：component-scan base-package =com.cinebot/> 
< mvc：annotation-driven /> 
< bean id =sessionFactoryclass =org.springframework.orm.hibernate4.LocalSessionFactoryBean> 
< property name =dataSourceref =dataSource/> 
< property name =packagesToScan> 
< list> 
<值> com.cinebot.db.entity< /值> 
< / list> 
< / property> 
< property name =hibernateProperties> 
<道具> 
< prop key =hibernate.dialect> org.hibernate.dialect.HSQLDialect< / prop> 
< prop key =hibernate.show_sql> true< / prop> 
< /道具> 
< / property> 
< / bean> 
 
< bean id =transactionManagerclass =org.springframework.orm.hibernate4.HibernateTransactionManager> 
< property name =sessionFactoryref =sessionFactory/> 
< / bean> 
 
< tx：annotation-driven /> 
  

这是错误的代码：

  @Transactional 
 public static< T> T get（Class< T>类，序列化id）抛出异常{
 if（id == null）return null; 
 T obj =（T）HibernateUtil.getSessionFactory（）。getCurrentSession（）。get（classe，id）; 
 return obj; 
} 
  

这是一个例外：

  org.hibernate.HibernateException：如果没有活动事务，则get无效
  

这是一个实例：

  package com.cinebot.db.entity; 
 
 //生成3-lug-2012 10.31.04 by Hibernate Tools 3.4.0.CR1 
 
 import javax.persistence.Column; 
 import javax.persistence.Entity; 
 import javax.persistence.Id; 
 import javax.persistence.Table; 
 $ b $ ** 
 *由hbm2java生成的Dettagli 
 * / 
 @Entity 
 @Table（name =dettagli，catalog =cinebot ）
 public class Dettagli实现java.io.Serializable {
 
 private static final long serialVersionUID = 1L; 
私人字符串nome; 
私有字符串valore; 
 $ b $ public Dettagli（）{
} 
 
 public Dettagli（String nome）{
 this.nome = nome; 
} 
 
 public Dettagli（String nome，string valore）{
 this.nome = nome; 
 this.valore = valore; 
 
 $ b $ @ @ 
 @Column（name =nome，unique = true，nullable = false，length = 32）
 public String getNome（）{ 
返回this.nome; 
} 
 
 public void setNome（String nome）{
 this.nome = nome; 
 
 
 @Column（name =valore，length = 65535）
 public String getValore（）{
 return this.valore; 
} 
 
 public void setValore（String valore）{
 this.valore = valore; 
} 
 
} 
  

我不明白为什么不是足够的@Transactional注释来自动打开事务。我错过了什么？

谢谢

解决方案

可以重新检查以下几点（尽管我不会推荐使用相同的）。

• 您需要在控制器中注入/ autowire spring DAO。使用@Repository注释DAO


• 确保您的spring扫描并为由context：component-scan完成的DAO类创建bean。这里你的dao类应该在给定的包/子包内。

我建议在Controller和DAO之间使用服务层。将服务方法注释为@transactional，它在DAO中调用方法。请记住，您不应创建任何bean的新实例，以调用其中的方法，而是注入/ autowire Service-> Controller和DAO-> Service。

i have a trouble opening Hibernate transaction.

This is the configuration:

    <context:annotation-config />
    <context:component-scan base-package="com.cinebot" />
    <mvc:annotation-driven />
    <bean id="sessionFactory" class="org.springframework.orm.hibernate4.LocalSessionFactoryBean">
        <property name="dataSource" ref="dataSource" />
        <property name="packagesToScan">
            <list>
                <value>com.cinebot.db.entity</value>
            </list>
        </property>
        <property name="hibernateProperties">
            <props>
                <prop key="hibernate.dialect">org.hibernate.dialect.HSQLDialect</prop>
                <prop key="hibernate.show_sql">true</prop>
            </props>
        </property>
    </bean>

    <bean id="transactionManager" class="org.springframework.orm.hibernate4.HibernateTransactionManager">
        <property name="sessionFactory" ref="sessionFactory" />
    </bean>

<tx:annotation-driven />

And this is the faulty code:

@Transactional
public static <T> T get(Class<T> classe, Serializable id) throws Exception {
    if(id==null) return null;
    T obj = (T) HibernateUtil.getSessionFactory().getCurrentSession().get(classe, id);
    return obj;
}

This is the exception:

org.hibernate.HibernateException: get is not valid without active transaction

This is an example entitiy:

package com.cinebot.db.entity;

// Generated 3-lug-2012 10.31.04 by Hibernate Tools 3.4.0.CR1

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;

/**
 * Dettagli generated by hbm2java
 */
@Entity
@Table(name = "dettagli", catalog = "cinebot")
public class Dettagli implements java.io.Serializable {

    private static final long serialVersionUID = 1L;
    private String nome;
    private String valore;

    public Dettagli() {
    }

    public Dettagli(String nome) {
        this.nome = nome;
    }

    public Dettagli(String nome, String valore) {
        this.nome = nome;
        this.valore = valore;
    }

    @Id
    @Column(name = "nome", unique = true, nullable = false, length = 32)
    public String getNome() {
        return this.nome;
    }

    public void setNome(String nome) {
        this.nome = nome;
    }

    @Column(name = "valore", length = 65535)
    public String getValore() {
        return this.valore;
    }

    public void setValore(String valore) {
        this.valore = valore;
    }

}

I can not understand why is not enought the @Transactional annotation to auto-open the transaction. What i missed?

Thanks

解决方案

Based on the comments You can recheck below points(though I would not recommend using same)

• You need to inject/autowire spring DAO into your controller. Annotate the DAO with @Repository
• Make sure your spring scans and create bean for your DAO class which is done by context:component-scan. Here your dao class should be inside the given package/subpackage.

I would recommend to use service layer between your Controller and DAO. Annotate the service method as @transactional which calls method in DAO. Remember you should not create new instance of any bean in order to call methods in it but inject/autowire Service->Controller and DAO->Service.

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