Hibernate @Filter枚举集合

gre*_*red 6 java postgresql enums hibernate

我需要弄清楚如何将基于注释的过滤应用于枚举的参数列表,定义如下:

@Column(name = "target_status")
@Enumerated(EnumType.STRING)
@Type(type="PGEnumConverter", parameters = {
    @Parameter(name = "enumClassName",
               value = "com.company.model.campaign.CampaignTarget$Status")
})
private Status targetStatus;
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所以我@FilterDef看起来像这样:

    @FilterDef(name="filterCampaignTargetByStatuses",
               defaultCondition="target_status in (:statuses)",
               parameters = @ParamDef(name = "statuses", type = "string"))
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当我启用过滤器时,它看起来像这样:

    session.enableFilter("filterCampaignTargetByStatuses").
    setParameterList("statuses", statuses);
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我从hibernate获得的错误是:

 org.hibernate.HibernateException: Incorrect type for parameter [statuses]
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数据在PostgreSQL中并且类型的定义如下:

CREATE TYPE statuscmp AS ENUM ('ACTIVE','INACTIVE','PAUSED','DRAFT','SCHEDULED','ENDED','ARCHIVED');
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我已经看到很多关于如何针对单个Enum值进行标准查询和过滤的SO问题,但是还没有关于对一 Enum值进行过滤的问题.有没有办法明确地投射个别价值观?

Kev*_*ers 1

一般来说,您不必“转换”该值,事实上您只需以存储的形式传递值即可。

如果我们假设您的字段仅被注释为该@Enumerated(EnumType.STRING)列将是一个普通的 varchar 字段。(将 java 类型映射到 postgres 枚举是另一个大主题。)

如果您现在想要将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,toString()如果它是 java ,则调用它的方法enum

例如,这是你的枚举:

public enum EntityStatus {
    A, B, C;
}
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这是您的实体:

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;
    }

}
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这可能是您要过滤的代码:

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();
}
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或者Java 8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();
}
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因此,仅过滤值的集合是可能的。