Question

这是我的dao方法，我尝试从数据库中获取用户详细信息：

public UserInfo findUserInfo(String userName) {
    String sql = "Select new " + UserInfo.class.getName() + "(u.username,u.password) "//
            + " from " + User.class.getName() + " u where u.username = :username ";

    Session session = sessionFactory.getCurrentSession();

    Query query = session.createQuery(sql);
    query.setParameter("username", userName);

    return (UserInfo) query.uniqueResult();
}

当我尝试执行它时，我收到此错误：

Caused by: org.hibernate.QueryParameterException: could not locate named parameter [username]
    at org.hibernate.engine.query.spi.ParameterMetadata.getNamedParameterDescriptor(ParameterMetadata.java:148)
    at org.hibernate.engine.query.spi.ParameterMetadata.getNamedParameterExpectedType(ParameterMetadata.java:165)
    at org.hibernate.internal.AbstractQueryImpl.determineType(AbstractQueryImpl.java:523)
    at org.hibernate.internal.AbstractQueryImpl.setParameter(AbstractQueryImpl.java:493)
    at org.spajic.stefan.springhibernatesecurity.dao.UserInfoDAO.findUserInfo(UserInfoDAO.java:33)
    at org.spajic.stefan.springhibernatesecurity.dao.UserInfoDAO$$FastClassBySpringCGLIB$$c14aad39.invoke(<generated>)
    at org.springframework.cglib.proxy.MethodProxy.invoke(MethodProxy.java:204)
    at org.springframework.aop.framework.CglibAopProxy$CglibMethodInvocation.invokeJoinpoint(CglibAopProxy.java:717)
    at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:157)
    at org.springframework.transaction.interceptor.TransactionInterceptor$1.proceedWithInvocation(TransactionInterceptor.java:99)
    at org.springframework.transaction.interceptor.TransactionAspectSupport.invokeWithinTransaction(TransactionAspectSupport.java:281)
    at org.springframework.transaction.interceptor.TransactionInterceptor.invoke(TransactionInterceptor.java:96)
    at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:179)
    at org.springframework.aop.framework.CglibAopProxy$DynamicAdvisedInterceptor.intercept(CglibAopProxy.java:653)
    at org.spajic.stefan.springhibernatesecurity.dao.UserInfoDAO$$EnhancerBySpringCGLIB$$3fdca957.findUserInfo(<generated>)
    at org.spajic.stefan.springhibernatesecurity.authentication.MyUserDetailsService.loadUserByUsername(MyUserDetailsService.java:25)
    at org.springframework.security.authentication.dao.DaoAuthenticationProvider.retrieveUser(DaoAuthenticationProvider.java:114)
    ... 36 more

这里也是User类和UserInfo类。用户是hibernate实体，UserInfo是我使用的模型。

@Entity
@Table(name = "Users")
public class User {

    private String username;
    private String password;
    private boolean enabled;

    @Id
    @Column(name = "username", length = 36, nullable = false)
    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    @Column(name = "password", nullable = false)
    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    @Column(name = "enabled", nullable = false)
    public boolean isEnabled() {
        return enabled;
    }

    public void setEnabled(boolean enabled) {
        this.enabled = enabled;
    }

}

public class UserInfo {

   private String userName;
   private String password;

   public UserInfo()  {

   }

   // Do not change this constructor, it used in hibernate Query.
   public UserInfo(String userName, String password) {
       this.userName = userName;
       this.password = password;
   }

   public String getUserName() {
       return userName;
   }

   public void setUserName(String userName) {
       this.userName = userName;
   }

   public String getPassword() {
       return password;
   }

   public void setPassword(String password) {
       this.password = password;
   }

}

任何帮助？

Answer 1

看起来该属性在您的POJO中被命名为UserInfo：userName，并且您正在尝试使用用户名

我的意思是你使用的是小'n'而不是'N'

Answer 2

希望这有帮助，查询甚至接受位置参数：

也许您可以在查询中尝试u.username = ?1 和query.setParameter(1, userName);设置参数

无论如何，您可以将结果发送到用户对象并使用它，我认为您不需要 UserInfo 。

无法找到命名参数Hibernate

2 个答案: