Question

我有JPA实体，如下所示： QueryDSL JPA syntax error with contains on Set?

现在，我尝试在单个查询中对Set tags进行多项限制：

Set<Tag> withTags = ...;
Set<Tag> withoutTags = ...;

q.where(license.tags.any().in(withTags));
q.where(license.tags.any().in(withoutTags).not());

执行查询时，我得到以下异常：

Exception [EclipseLink-8019] (Eclipse Persistence Services - 2.0.1.v20100213-r6600): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Error compiling the query [select distinct license
from License license
where exists (select license_tags
from Tag license_tags
where license_tags member of license.tags and license_tags = ?1)
and not exists (select license_tags
from Tag license_tags
where license_tags member of license.tags and license_tags = ?2)]
multiple declaration of identification variable [license_tags], previously declared as [Tag license_tags].

我尝试将as("withTags")插入到查询中，但我可以执行的位置是在any()之后，它将JP中的AS插入错误的地方，关于我正在尝试解决的重复问题。我可以在tags之后插入它，但后来我得到一个SimpleExpression作为返回，我无法执行any()。

还有其他想法如何防止这种重复的识别变量？

此外，上述声明仅在给定Set withTags / withoutTags仅包含单个值时才有效。如果存在多个值，则抛出以下异常：

Exception [EclipseLink-6075] (Eclipse Persistence Services - 2.0.1.v20100213-r6600): org.eclipse.persistence.exceptions.QueryException
Exception Description: Object comparisons can only use the equal() or notEqual() operators.  Other comparisons must be done through query keys or direct attribute level comparisons. 
Expression: [Relation operator  IN   Base my.package.Tag   Parameter 1]
select distinct license
from License license
where exists (select license_tags
from Tag license_tags
where license_tags member of license.tags and license_tags in ?1)
    at org.eclipse.persistence.exceptions.QueryException.invalidOperatorForObjectComparison(QueryException.java:614)
    at org.eclipse.persistence.internal.expressions.RelationExpression.normalize(RelationExpression.java:393)
    at org.eclipse.persistence.internal.expressions.CompoundExpression.normalize(CompoundExpression.java:226)
    at org.eclipse.persistence.internal.expressions.CompoundExpression.normalize(CompoundExpression.java:218)
    at org.eclipse.persistence.internal.expressions.SQLSelectStatement.normalize(SQLSelectStatement.java:1306)
    at org.eclipse.persistence.internal.expressions.SubSelectExpression.normalizeSubSelect(SubSelectExpression.java:134)
    at org.eclipse.persistence.internal.expressions.ExpressionNormalizer.normalizeSubSelects(ExpressionNormalizer.java:93)
    at org.eclipse.persistence.internal.expressions.SQLSelectStatement.normalize(SQLSelectStatement.java:1379)
    at org.eclipse.persistence.internal.queries.ExpressionQueryMechanism.buildNormalSelectStatement(ExpressionQueryMechanism.java:482)
    at org.eclipse.persistence.internal.queries.ExpressionQueryMechanism.prepareSelectAllRows(ExpressionQueryMechanism.java:1553)
    at org.eclipse.persistence.queries.ReadAllQuery.prepareSelectAllRows(ReadAllQuery.java:793)
    at org.eclipse.persistence.queries.ReadAllQuery.prepare(ReadAllQuery.java:734)
    at org.eclipse.persistence.queries.DatabaseQuery.checkPrepare(DatabaseQuery.java:464)
    at org.eclipse.persistence.queries.ObjectLevelReadQuery.checkPrepare(ObjectLevelReadQuery.java:732)
    at org.eclipse.persistence.queries.DatabaseQuery.prepareCall(DatabaseQuery.java:1577)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:240)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:173)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:125)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:109)
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1326)
    at sun.reflect.GeneratedMethodAccessor552.invoke(Unknown Source)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    at java.lang.reflect.Method.invoke(Method.java:597)
    at org.jboss.weld.util.reflection.SecureReflections$13.work(SecureReflections.java:304)
    at org.jboss.weld.util.reflection.SecureReflectionAccess.run(SecureReflectionAccess.java:54)
    at org.jboss.weld.util.reflection.SecureReflectionAccess.runAsInvocation(SecureReflectionAccess.java:163)
    at org.jboss.weld.util.reflection.SecureReflections.invoke(SecureReflections.java:298)
    at org.jboss.weld.bean.proxy.ClientProxyMethodHandler.invoke(ClientProxyMethodHandler.java:113)
    at org.jboss.weld.util.CleanableMethodHandler.invoke(CleanableMethodHandler.java:43)
    at javax.persistence.EntityManager_$$_javassist_131.createQuery(EntityManager_$$_javassist_131.java)
    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)

使用EclipseLink 2.4

Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.4.0.v20120608-r11652): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: org.postgresql.util.PSQLException: The SQL datatype to be used for an instance of mypackage.Tag cannot be determined. Use 'setObject()' with an explizit type, to define it.
Error Code: 0
Call: SELECT DISTINCT t0.ID, ...all the other properties...
FROM LICENSE t0, WHERE ((NOT EXISTS (SELECT ? FROM LicenseTags t5, TAG t4, TAG t3 WHERE (((t3.ID = t4.ID) AND (t3.ID IN (?,?))) AND ((t5.License_ID = t0.ID) AND (t4.ID = t5.tags_ID))))))

目前我尝试使用以下QueryDSL语法解决此问题：

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

for (Tag tag : withoutTags) {
    q.where(license.tags.contains(tag).not());
}

前一部分确实像魅力一样，但后者并没有返回预期的结果。 withoutTags中包含标记的许可证不会从结果集中排除。

后一语句的JPQL和SQL如下所示：

select distinct license
from License license
where not ?1 member of license.tags

SELECT DISTINCT t1.ID, ...all the other properties...
FROM LicenseTags t2, LICENSE t1, TAG t0
WHERE (NOT (133170 = t0.ID) AND (t2.License_ID = t1.ID) AND (t0.ID = t2.tags_ID))

JPQL对我来说很好看，但如果许可证有多个与之关联的标记，那么SQL显然会失败。所以我认为这实际上是EclipseLink的翻译失败的情况。如果这是我正在使用的版本的已知错误，我会看看。这篇论文得到JPQL "NOT MEMBER OF" query using criteria API的支持，但在这种情况下，问题只发生在使用creteria api而不是JPQL时。这个错误的翻译仍然存在于EclipseLink 2.4 RC 2中。最后，这是一个解决方法，它为“无标记”部分实现了它的意义：

Collection<Integer> tagIds = new ArrayList<Integer>();
for (Tag tag : withoutTags) {
    tagIds.add(tag.getId());
}
q.where(license.tags.any().id.in(tagIds).not());

此致，蒂尔曼

Answer 1

您可以使用联接而不是子选择吗？这也会更有效率。

您也可以尝试EclipseLink 2.4，它可能没有这些问题。

QueryDSL支持连接，包括innerJoin（）和leftJoin（），你应该使用它。

Answer 2

目前EclipseLink和QueryDSL中存在拒绝使用直接语句的错误：

q.where(license.tags.any().in(withTags));
q.where(license.tags.any().in(withoutTags).not());

而是必须使用解决方法：

Collection<Integer> withTagIds = new ArrayList<Integer>();
for (Tag tag : withTags) {
    withTagIds.add(tag.getId());
}
q.where(license.tags.any().id.in(withTagIds));

Collection<Integer> withoutTagIds = new ArrayList<Integer>();
for (Tag tag : withoutTags) {
    withoutTagIds .add(tag.getId());
}
q.where(license.tags.any().id.in(withoutTagIds ).not());

QueryDSL重复标识变量/等于语法错误与任何on？

2 个答案: