Hibernate@filter枚举集合

Question

我需要弄清楚如何使用定义为以下内容的枚举参数列表应用基于注释的筛选：

@Column(name = "target_status")
@Enumerated(EnumType.STRING)
@Type(type="PGEnumConverter", parameters = {
    @Parameter(name = "enumClassName",
               value = "com.company.model.campaign.CampaignTarget$Status")
})
private Status targetStatus;

所以我的@filterdef如下所示：

    @FilterDef(name="filterCampaignTargetByStatuses",
               defaultCondition="target_status in (:statuses)",
               parameters = @ParamDef(name = "statuses", type = "string"))

    session.enableFilter("filterCampaignTargetByStatuses").
    setParameterList("statuses", statuses);

 org.hibernate.HibernateException: Incorrect type for parameter [statuses]

CREATE TYPE statuscmp AS ENUM ('ACTIVE','INACTIVE','PAUSED','DRAFT','SCHEDULED','ENDED','ARCHIVED');

我见过很多关于如何对单个枚举值进行条件查询和筛选的问题，但还没有关于对一组枚举值进行筛选的问题。是否有一种方法可以显式地强制转换单个值？

Answer 1

通常不必“强制转换”值，实际上只需以存储形式传递值即可。

如果我们假设您的字段只注释为@enumerated(enumtype.string)，则该列将是一个普通的varchar字段。（将java类型映射到postgres枚举是另一个重要的主题。）

如果您现在想将status枚举实例列表与数据库中的相关字符串值进行比较，请将其作为字符串集合传递，换句话说，如果它是javaenum则调用它的tostring()方法。

public enum EntityStatus {
    A, B, C;
}

这是你的实体：

import javax.persistence.Entity;
import javax.persistence.EnumType;
import javax.persistence.Enumerated;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;

import org.hibernate.annotations.Filter;
import org.hibernate.annotations.FilterDef;
import org.hibernate.annotations.ParamDef;

@Entity
@FilterDef(name = "byMultipleStates", defaultCondition = "status in (:states)", parameters = @ParamDef(name = "states", type = "string"))
@Filter(name = "byMultipleStates", condition = "status in (:states)")
public class StatusEntity {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long id;

    @Enumerated(EnumType.STRING)
    private EntityStatus status;

    public long getId() {
        return id;
    }

    public EntityStatus getStatus() {
        return status;
    }

    public void setStatus(EntityStatus status) {
        this.status = status;
    }

}

这可能是您要筛选的代码：

public List<StatusEntity> filterByStates(final Set<EntityStatus> states) {
    final Session hibernateSession = entityManager.unwrap(Session.class);
    hibernateSession.enableFilter("byMultipleStates").setParameterList("states",
            states.stream().map(state -> state.toString()).collect(Collectors.toList()));
    final Query query = hibernateSession.createQuery("SELECT e FROM StatusEntity e");

    return query.list();
}

或者以Java8之前的方式：

public List<StatusEntity> filterByStates(final Set<EntityStatus> states) {
    final Set<String> statesAsString = new HashSet<>();
    for (final EntityStatus state : states) {
        statesAsString.add(state.toString());
    }

    final Session hibernateSession = entityManager.unwrap(Session.class);
    hibernateSession.enableFilter("byMultipleStates").setParameterList("states", statesAsString);
    final Query query = hibernateSession.createQuery("SELECT e FROM StatusEntity e");

    return query.list();
}

因此，只对值集合进行筛选是可能的。