休眠@Embeddable类为未引用的列抛出"SQLServerException: Invalid column name"

Question

我一直为这件事心烦意乱。我有一个类像这样拉动对象:

public UserDTO getUser(String login)
{
    String jql = "select entity from User as entity  where entity.abcUserId = :userID ";
    Query query = _entityManager.createQuery(jql)
            .setParameter("userID", login);
    try
    {
        Object user = query.getSingleResult();
        return ((User) user).extractObject();
    }
    catch(NoResultException e)
    {
        LOG.error(e.getMessage());
        return null;
    }
}

与它正在拉的类:

@Entity
@Table(name="ABC_User")
public class User implements Serializable,EntityObject<UserDTO>
{
private static final long   serialVersionUID    = 1L;
private Long id;
private String abcUserId;
private Company company;
private List<UserPreference> userPreferences;
@Id
@GeneratedValue
public Long getId()
{
    return id;
}
public void setId(Long id)
{
    this.id = id;
}
@ManyToOne(fetch=FetchType.LAZY)
@JoinColumn(name="Company_ID", nullable=false)
public Company getCompany()
{
    return company;
}
public void setCompany(Company customer)
{
    this.company = customer;
}
/**
 * @return the userPreference
 */
@ElementCollection(fetch=FetchType.EAGER)
@JoinTable(name="ABC_USER_PREFERENCES")
@Cascade(value={org.hibernate.annotations.CascadeType.ALL})
public List<UserPreference> getUserPreferences()
{
    if(userPreferences == null)
    {
        userPreferences = new ArrayList<UserPreference>();
    }
    return userPreferences;
}
/**
 * @param userPreference the userPreference to set
 */
public void setUserPreferences(List<UserPreference> userPreferences)
{
    this.userPreferences = userPreferences;
}
/**
 * @return the abcUserId
 */
@Column(name="ABC_USER_ID", length=10, nullable=false)
public String getAbcUserId()
{
    return abcUserId;
}
/**
 * @param abcUserId the abcUserId to set
 */
public void setAbcUserId(String abcUserId)
{
    this.abcUserId = abcUserId;
}

引用这个可嵌入对象:

@Embeddable
public class UserPreference implements Serializable
{
private static final long   serialVersionUID    = 1L;
private String prefKey;
private String prefValue;
public UserPreference() {}
public UserPreference(String key, String value)
{
    this.prefKey = key;
    this.prefValue = value;
}
@Column(nullable=false, length=255)
public String getPrefKey()
{
    return prefKey;
}
public void setPrefKey(String key)
{
    this.prefKey = key;
}
@Column(nullable=false, length=1048576)
public String getPrefValue()
{
    return prefValue;
}
public void setPrefValue(String value)
{
    this.prefValue = value;
}
}

所以，冗长的代码块被删除了一些东西，这是行不通的。每次我试图从数据库中拉一个用户，它抛出"SQLServerException:无效的列名'User_id'"。User_id从未在我的项目中引用(我已经检查过)，它的总是 abc_user。我可以看到对象在Eclipse调试器中被拉到一起，它达到了添加UserPreferences列表的地步，然后解体了。如果我注释掉User类的UserPreferences部分，它就会成功拉出(并在使用它们的其他地方中断)。

我错过了什么?

Answer 1

参见以下代码部分

@Column(name="ABC_USER_ID", length=10, nullable=false)
public String getAbcUserId()
{
    return abcUserId;
}

检查表是否有列名'ABC_USER_ID'。然后，尝试将此代码放在@JoinTable部分

Answer 2

尝试为User.id定义@Column。它有一个@Id，但没有@Column。可能是Hibernate假设该字段的列是_id，结果是'User_id'，当它试图将返回的数据映射到对象时找不到它。