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我尝试将 JPA 与 GAE 一起使用，但我遇到了下一个问题：

当我执行此查询时：

select x From Advert x where x.advertType in (15)

或者

select x From Advert x where x.advertType in (20)

实体管理器在两种情况下都返回一个实体，但是当我尝试执行时

select x From Advert x where x.advertType in (15,20)

返回零实体

这是我的映射：

@Entity
public class Advert extends DomainObject implements Comparable<Advert> {
    @ManyToOne(cascade = CascadeType.PERSIST, fetch = FetchType.EAGER)
    @Unowned
    private AdvertType advertType;
    ...
}

@Entity
public class AdvertType extends DomainObject implements Comparable<AdvertType> {
    private Integer position;
    private String typeId;
    private String titleMain;
}

@Entity
@MappedSuperclass
public abstract class DomainObject implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Key key;
}

日志：

2013-04-20 12:48:17 DEBUG Query:58 - QueryCompilation:
  [from:ClassExpression(alias=x)]
  [filter:DyadicExpression{DyadicExpression{PrimaryExpression{x.advertType}  =  Literal{15}}  OR  DyadicExpression{PrimaryExpression{x.advertType}  =  Literal{20}}}]
  [symbols: x type=net.org.selector.animals.domain.Advert]
2013-04-20 12:48:17 DEBUG Query:58 - JPQL Query : Executing "SELECT x FROM Advert x WHERE x.advertType in (15,20)" ...
2013-04-20 12:48:17 DEBUG Query:58 - Query compiled as : Kind=net.org.selector.animals.domain.Advert Filter : advertType_key=AdvertType(15) AND advertType_key=AdvertType(20) [QUERY-TYPE=NORMAL]
2013-04-20 12:48:17 DEBUG Native:58 - Executing query in datastore for SELECT x FROM Advert x WHERE x.advertType in (15,20)

那么为什么查询编译为：

advertType_key=AdvertType(15) AND advertType_key=AdvertType(20)

代替

advertType_key=AdvertType(15)或advertType_key=AdvertType(20)

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1 回答 1

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您的查询有缺陷。“x.advertType”是 AdvertType 类型，因此它不能“在”数字列表中。如果“x.advertType.typeId”在字符串列表中，则可能会使用它。日志还告诉您实际调用了哪些低级查询

于 2013-04-20T07:12:44.160 回答