java – Hibernate @MapKeyColumn和表继承导致Unknown列类型异常

我从hibernate升级4.2.5.Final到4.3.6.Final,4.3.6 hibernate libs导致 mysql未知列类型异常.以下课程已被简化,因为我无法全面显示我的公司生产代码.

@Entity
@Table(name = "area")
public class Area {
    private Integer id;
    private Map<BasicType,BasicConfiguration> configurationsMap =
        new HashMap<BasicType,BasicConfiguration>();

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    @OnetoMany(fetch = FetchType.EAGER,cascade = {CascadeType.ALL},orphanRemoval = true)
    @JoinTable(name = "area_configuration",joinColumns = {@JoinColumn(name = "area_id")},inverseJoinColumns = {@JoinColumn(name = "basic_configuration_id")})
    @MapKeyEnumerated(EnumType.STRING)
    @MapKeyColumn(name = "type")
    public Map<BasicType,BasicConfiguration> getConfigurationsMap () {
        return configurationsMap;
    }

BasicType只是一个枚举

public enum BasicType {
    TYPE1,TYPE2,TYPE3,TYPE4,TYPE5;
}

而基本配置是：

@Entity
@Table(name = "basic_configuration")
@Inheritance(strategy = InheritanceType.JOINED)
@discriminatorColumn(name = "type",discriminatorType = discriminatorType.STRING)
public abstract class BasicConfiguration {

    private Integer id;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }
}

我有一个测试,试图持续一个区域对象到一个MysqL数据库,产生以下错误：

**Caused by: com.MysqL.jdbc.exceptions.jdbc4.MysqLSyntaxErrorException: UnkNown column 'type' in 'field list'**
    at sun.reflect.GeneratedConstructorAccessor73.newInstance(UnkNown Source)
    at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)

    at java.lang.reflect.Constructor.newInstance(Constructor.java:526)
    at com.MysqL.jdbc.Util.handleNewInstance(Util.java:408)
    at com.MysqL.jd

生成的hibernate代码显示它正在尝试将类型值插入到basic_configuration表中,而不是area_configuration表中：

**Hibernate: insert into basic_configuration (entity_version,type) values (?,TYPE1)**
Tests run: 9,Failures: 0,Errors: 9,Skipped: 0,Time elapsed: 0.208 sec <<< FAILURE! - 
javax.persistence.PersistenceException: org.hibernate.exception.sqlGrammarException: Could not execute statement
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1387)
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1310)
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1316)
at org.hibernate.ejb.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:881)
at sun.reflect.Ge

这个错误似乎已经在hibernate 4.2.9中引入了.Final起始,4.2.9以下的版本.Final似乎没有这个问题,有人知道我该如何解决这个问题？非常感谢.

解决方法

我记得有一个类似的问题.在这里查看你的代码：

@Entity
@Table(name = "basic_configuration")
@Inheritance(strategy = InheritanceType.JOINED)
@discriminatorColumn(name = "type",discriminatorType = discriminatorType.STRING)
public abstract class BasicConfiguration {

    private Integer id;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }
}

在这一行 – >

@discriminatorColumn(name = "type",discriminatorType = discriminatorType.STRING)

您正在使用sql保留字,“like”.查看此列表：

Reserved words