关于java：QueryDSL JPA语法错误，包含在Set上？

QueryDSL JPA syntax error with contains on Set?

我有一个类似于以下内容的JPA实体bean：

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@Entity
class License {
@ManyToMany(fetch = FetchType.EAGER)
@JoinTable(name ="LicenseTags")
Set<Tag> tags;

// Skipped remaining members
}

Tag本身也是具有ID和名称的Entity。现在，我要查询已附加某些标签的许可证。当我尝试以下查询时

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Set<Tag> tags = ...;
final QLicense license = QLicense.license;
JPAQuery q = new JPAQuery(entityManager).from(license);

for (Tag tag : tags) {
q.where(license.tags.contains(tag));
}

Collection<License> result = q.listDistinct(license);

我在listDistinct

的行上收到以下异常

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java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager:
Exception Description: Syntax error parsing the query [select distinct license
from License license
where ?1 in elements(license.tags)]: unexpected token [in].
Internal Exception: NoViableAltException(35!=[685:1: inExpression[boolean not, Object left] returns [Object node] : (t= IN n= inputParameter | t= IN LEFT_ROUND_BRACKET (itemNode= inItem ( COMMA itemNode= inItem )* | subqueryNode= subquery ) RIGHT_ROUND_BRACKET );])
at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1328)
at com.sun.enterprise.container.common.impl.EntityManagerWrapper.createQuery(EntityManagerWrapper.java:425)
at com.mysema.query.jpa.impl.DefaultSessionHolder.createQuery(DefaultSessionHolder.java:35)
at com.mysema.query.jpa.impl.AbstractJPAQuery.createQuery(AbstractJPAQuery.java:139)
at com.mysema.query.jpa.impl.AbstractJPAQuery.createQuery(AbstractJPAQuery.java:108)
at com.mysema.query.jpa.impl.AbstractJPAQuery.list(AbstractJPAQuery.java:276)
at com.mysema.query.support.ProjectableQuery.listDistinct(ProjectableQuery.java:104)

从解析器异常输出中，我只能猜测也许缺少括号。

我在查询集合中包含的值时做错了吗？

我正在使用GlassFish Server开源版本3.0.1(内部版本22)，而该版本又使用EclipseLink捆绑版本：2.0.1.v20100213-r6600

关于蒂尔曼，