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分享hibernate4 使用getCurrentSession().createQuery报错，求助
public List findByProperty(String propertyName, Object value) {
log.debug("finding FmUser instance with property: " + propertyName
+ ", value: " + value);
try {
String queryString = "from FmUser as model where model."
+ propertyName + "= ?";
Query queryObject = getCurrentSession().createQuery(queryString);
queryObject.setParameter(0, value);
return queryObject.list();
} catch (RuntimeException re) {
log.error("find by property name failed", re);
throw re;
}
}调试到“getCurrentSession().createQuery”报一个错误：
Source not found for FmUserDAO$$FastClassByCGLIB$$deecd5cd.invoke(int, Object, Object[]) line: not available
提示找不到FmUser，但我在com.fundclien.entity.FmUser是存在的，它的代码是：
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Table;
import javax.persistence.UniqueConstraint;
/**
* FmUser entity. @author MyEclipse Persistence Tools
*/
@Entity
@Table(name = "FM_USER", schema = "GF", uniqueConstraints = @UniqueConstraint(columnNames = "VC_USER_CODE"))
public class FmUser implements java.io.Serializable {
@Id
@GeneratedValue
@Column(name = "L_USER_ID", unique = true, nullable = false, precision = 10, scale = 0)
public Long getLUserId() {
return this.LUserId;
}
public void setLUserId(Long LUserId) {
this.LUserId = LUserId;
}
@Column(name = "VC_USER_CODE", unique = true, nullable = false, length = 16)
public String getVcUserCode() {
return this.vcUserCode;
}
}
下面是applicationContext.xml配置：
<?xml version="1.0" encoding="UTF-8"?>
<beans
xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:p="http://www.springframework.org/schema/p"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx.xsd" xmlns:tx="http://www.springframework.org/schema/tx">
<bean id="dataSource"
class="org.apache.commons.dbcp.BasicDataSource">
<property name="driverClassName" value="oracle.jdbc.driver.OracleDriver"></property>
<property name="url"
value="jdbc:oracle:thin:@127.0.0.1:1521:orcl">
</property>
<property name="username" value="gf"></property>
<property name="password" value="gf"></property>
</bean>
<bean id="sessionFactory"
class="org.springframework.orm.hibernate4.LocalSessionFactoryBean">
<property name="dataSource">
<ref bean="dataSource" />
</property>
<property name="hibernateProperties">
<props>
<prop key="hibernate.dialect">
org.hibernate.dialect.Oracle9Dialect
</prop>
<prop key="hbm2ddl.auto">update</prop>
<prop key="hibernate.show_sql">true</prop>
</props>
</property>
<property name="annotatedClasses">
<list>
<value>com.fundclien.entity.FmUser</value>
</list>
</property>
</bean>
<bean id="transactionManager"
class="org.springframework.orm.hibernate4.HibernateTransactionManager">
<property name="sessionFactory" ref="sessionFactory" />
</bean>
<tx:annotation-driven transaction-manager="transactionManager" />
<bean
id="fmUserDao" class="com.fundclien.dao.FmUserDAO">
<property name="sessionFactory">
<ref bean="sessionFactory" />
</property>
</bean>
<bean id="login" class="com.fundclient.actions.loginAction">
<property name="fmUserDao" ref="fmUserDao" />
</bean>
</beans>
求前辈帮我看看是怎么回事啊？我调用FmUser instance = (FmUser) getCurrentSession().get("com.fundclien.entity.FmUser", id);则可以取得。
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Roy_Sashulin 2014-07-18
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:-) 我写错了，错误消息是No result defined for action com.fundclient.actions.loginAction and result error
suciver 2014-07-18
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action com.fundclient.actions.loginA这个action没有result的返回页面，楼主检查下这个action的配置
Roy_Sashulin 2014-07-18
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不行也，还是会报错，这是不是hibernate4的bug啊，或者是不是需要配置什么参数来技能呀
Roy_Sashulin 2014-07-18
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还有一个问题:
我在loginAction的execute函数执行完成后，在跳转页面时报了个错误：
HTTP Status 404 - No result defined for action com.fundclient.actions.loginA
public String execute() throws Exception{
boolean res = fmUserDao.validUser(userName, passWord);
if(res){
return SUCCESS;
}else{
return ERROR;
}
}
<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE struts PUBLIC "-//Apache Software Foundation//DTD Struts Configuration 2.1//EN" "http://struts.apache.org/dtds/struts-2.1.dtd">
<struts>
<constant name="struts.objectFactory" value="spring"></constant>
<package name="mypack" extends="struts-default">
<action name="index">
<result>/index.jsp</result>
</action>
<action name="login" class="com.fundclient.actions.LoginAction">
<result name="success">/index.jsp</result>
<result name="error">/login.jsp</result>
</action>
</package>
</struts>
whos2002110 2014-07-18
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你这样试一下：
String queryString = "from FmUser as model where model."
+ propertyName + "=:value";
Query queryObject = getCurrentSession().createQuery(queryString);
queryObject.setParameter("value", value);
Roy_Sashulin 2014-07-18
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感谢 lmj623565791
我已经按你指教的改了，果然好用啊。
to 版主大人 fangmingshijie
findByProperty(String propertyName, Object value) 是自动生成的，value是propertyName的值。
花谢尊前不敢香 2014-07-18
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queryObject.setParameter(0, value);
这个value应该是FmUser类型，通过返回去get 对应的propertyName。
鸿洋_ 2014-07-18
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如果你喜欢注解，把seesionFactory中的属性
<property name="annotatedClasses">
<list>
<value>com.fundclien.entity.FmUser</value>
</list>
</property>
改成
<property name="packagesToScan">
<value>com.fundclien.entity</value>
</property>
你也不嫌累，一个一个加~
Roy_Sashulin 2014-07-18
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我修改为字符串拼接就可以，如果用参数需要做什么配置吗？
String queryString = "from FmUser as model where model."
+ propertyName + "= '"+value+"'";
修改成这样就可以。
Roy_Sashulin 2014-07-18
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1.参数绑定这个还没有解决。
2.访问已经解决了
<action name="login" class="com.fundclient.actions.loginAction" method="execute">
改成
<action name="loginAction" class="com.fundclient.actions.loginAction" method="execute">
就好了。
