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2个POJO类扩展了一个基类,其中每个类映射到唯一的表。为什么Hibernate查询在字段列表错误中创建未知列“”?

hibernate-mapping inheritance hibernate mysql java

sofs1 Romain Manni-Bucau · 技术社区 · 6 年前

问题.java

@Entity
@Table(name="question")
@Access(value = AccessType.FIELD)
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public class Questions{

    private static final long serialVersionUID = 1L;

    private String qid;

    @Column(name="addedtime")
    private String addedtime;

    @Column(name="qlang")
    private String qlang;

    @Id
    @Column(name="qid")
    @GeneratedValue(generator = "uuid2")
    @GenericGenerator(name = "uuid2", strategy = "uuid2")
    @Access(value = AccessType.PROPERTY)
    public String getQid() {
        return qid;
    }

    public void setQid(String qid) {
        this.qid = qid;
    }

    @Access(value = AccessType.PROPERTY)
    public String getAddedtime() {
        return addedtime;
    }

    public void setAddedtime(String addedtime) {
        this.addedtime = addedtime;
    }

    @Access(value = AccessType.PROPERTY)
    public String getQlang() {
       return qlang;
    }

    public void setQlang(String qlang) {
       this.qlang = qlang;
    }
}

@Entity
@Table(name="MCQ")
@Access(value = AccessType.FIELD)
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public class MCQ extends Questions implements Serializable{

    @Column(name="option_1")
    @Access(value = AccessType.FIELD)
    private String option_1;

    @Access(value = AccessType.PROPERTY)
    public String getOption_1() {
        return option_1;
    }

    public void setOption_1(String option_1) {
        this.option_1 = option_1;
    }

    @Column(name="option_2")
    @Access(value = AccessType.FIELD)
    private String option_2;

    @Access(value = AccessType.PROPERTY)
    public String getOption_2() {
        return option_2;
    }

    public void setOption_2(String option_2) {
        this.option_2 = option_2;
    }

}

@Entity
@Table(name="TwoMarkQ")
@Access(value = AccessType.FIELD)
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public class TwoMarkQ extends Questions implements Serializable{

    @Column(name="option_1")
    @Access(value = AccessType.FIELD)
    private String option_1;

    @Access(value = AccessType.PROPERTY)
    public String getOption_1() {
        return option_1;
    }

    public void setOption_1(String option_1) {
        this.option_1 = option_1;
    }

    @Column(name="option_2")
    @Access(value = AccessType.FIELD)
    private String option_2;

    @Access(value = AccessType.PROPERTY)
    public String getOption_2() {
        return option_2;
    }

    public void setOption_2(String option_2) {
        this.option_2 = option_2;
    }

}

create table `question` (
  `qid` varchar(48) COLLATE utf8mb4_unicode_ci NOT NULL,
  `addedtime` datetime(6) NOT NULL DEFAULT CURRENT_TIMESTAMP(6),
  `qtype` varchar(32) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `qlang` varchar(48) COLLATE utf8mb4_unicode_ci NOT NULL,
   PRIMARY KEY (`qid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci

create table `MCQ`(
`qid` varchar(48) COLLATE utf8mb4_unicode_ci NOT NULL,
`option_1` bigint(20) DEGAULT `0`,
`option_2` bigint(20) DEGAULT `0`,
PRIMARY KEY (`qid`),
CONSTRAINT `mcq_ibfk1` FOREIGN KEY (`qid`) REFERENCES `question` (`qid`)
)ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci

create table `TwoMarkQ`(
`qid` varchar(48) COLLATE utf8mb4_unicode_ci NOT NULL,
`option_1` bigint(20) DEGAULT `0`,
`option_2` bigint(20) DEGAULT `0`,
PRIMARY KEY (`qid`),
CONSTRAINT `two_markq_ibfk1` FOREIGN KEY (`qid`) REFERENCES `question` (`qid`)
)ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci

Query query = session.createQuery("select q.qid, q.qtype from Questions q where q.qlang=:lang ORDER BY q.addedtime ASC");
            query.setParameter("lang", lang);
            query.setFirstResult(startingRow).setMaxResults(10);
            result = (List<Questions>) query.list();

result = (List<Questions>) query.list();

com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'addedtime' in 'field list'

请帮忙。在这上面打了3天。

PS:这是生成的查询

select questions0_.qid as col_0_0_, questions0_.qtype as col_1_0_ from ( select qid, addedtime, qlang, qtype, null as option_1, null as option_2 as class_ from MCQ union  select qid, addedtime, qlang, qtype, null as option_!, null as option_2 as class_ from TwoMarkQ)  questions0_ where questions0_.qlang=? order by questions0_.addedtime ASC limit ?

Query query = session.createQuery("select q.qid, q.qtype from Questions q where q.qlang=:lang ORDER BY q.addedtime ASC");

4 回复 | 直到 6 年前

vincent 6 年前

键入从服务器获取数据时使用的解决方案,但数据不反映当前模型。

vincent 6 年前

//DAO Generic
public List<T> findByCriteria(Criterion... criterion) throws DAOException {
        try {
            Criteria crit = getSession().createCriteria(getPersistentClass());
            for (Criterion c : criterion) {
                crit.add(c);
            }
            return crit.list();
        } catch (HibernateException ex) {
            logger.error(ex.getLocalizedMessage());
            throw new DAOException(ex);
        }
    }

//DAO person
    public List<Persona> findByCognome(String cognome) throws DAOException {
        try {
            Criteria criteria = getSession().createCriteria(Persona.class);
            criteria.add(Restrictions.eq("cognome", cognome).ignoreCase());
            criteria.addOrder(Order.asc("eta"));
            return criteria.list();
        } catch (DAOException ex) {
            throw new DAOException(ex.getLocalizedMessage());
        }
    }

Abhay Jain 6 年前

将继承策略更改为“已加入”。

您可能需要为问题表中的问题类型添加一列,该列是问题类型的鉴别器列,您需要将其作为问题类的注释放置。

 @Inheritance(strategy = InheritanceType.JOINED)
 @DiscriminatorColumn(name = "QUESTION_TYPE")
 public class Questions{
 .
 .
 }

@DiscriminatorValue("MCQ")
public class MCQ extends Questions implements Serializable{
.
}

@DiscriminatorValue("TwoMarkQ")
public class TwoMarkQ extends Questions implements Serializable{
.
}

sofs1 Romain Manni-Bucau 6 年前

Hibernate documentation

我们有3种选择。 InheritanceType.TABLE_PER_CLASS InheritanceType.SINGLE_TABLE InheritanceType.JOINED

JOINED: 绝对不是我想要的。所以这是一个被排除的选择。

SINGLE_TABLE:

我仍然有追随者

@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
@Polymorphism(type=PolymorphismType.EXPLICIT)

TABLE_PER_CLASS

“使用多态查询时,需要联合来获取基类表以及所有子类表。”—因此这也是一个排除的选项。