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我有此处概述的 JPA 实体： QueryDSL JPA 语法错误，包含在 Set 上？

现在我尝试Set tags在单个查询中设置多个限制：

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，出现以下异常：

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].

我试图插入as("withTags")查询，但我可以做的位置是在any()我试图解决的重复问题中将 JPQL 中的 AS 插入错误的位置。我可以在之后插入它，tags但随后我得到了一个SimpleExpression我无法执行的回报any()。

任何其他想法如何防止这种识别变量的重复？

此外，上面介绍的语句仅在给定的Set withTags/withoutTags仅包含单个值时才有效。如果存在多个值，则会引发以下异常：

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))))))

现在我尝试使用以下 QueryDSL 语法来解决这个问题：

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

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

前一部分确实像一个魅力，但后者没有返回预期的结果。带有标签的许可证withoutTags不会按原样从结果集中排除。

后一条语句的 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 对我来说看起来不错，但如果许可证有多个关联的标签，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。我会看看这是否是我正在使用的版本的已知错误。JPQL“NOT MEMBER OF”使用标准 API 查询在一定程度上支持本论文，但在这种情况下，问题仅在使用 creteria api 而不是 JPQL 时出现。这种错误的翻译在 EclipseLink 2.4 RC 2 中仍然存在。最后是一个解决方法，它对“没有标签”的部分意味着什么：

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

问候，蒂尔曼

4

2 回答 2

2

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，innerJoin() 和 leftJoin()，你应该使用它。

于 2012-06-19T14:58:51.043 回答
1

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());

相反，必须使用一种解决方法：

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());
于 2012-06-25T10:50:10.837 回答