QueryDSL重复识别变量/平等语法错误与任何设为？(QueryDSL duplicate ide

2019-07-30 00:23发布

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我有JPA实体这里概述： QueryDSL JPA语法错误与套装包括？

现在，我尝试对多重限制Set tags在一个查询： 

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

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

当执行查询，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].

我试图插入as("withTags")进入查询，但我能做的就是为以后的位置any()其插入在JPQL中的AS中关于复制的问题，我试图解决错误的地方。 我可以后插入tags ，但后来我得到一个SimpleExpression作为回报，我可以不执行any()

没有其他的想法如何复制标识变量是可以预防？

除了上面给出的语句，如果给定只做工作Set withTags / withoutTags只包含一个值。 如果有多个值出现以下异常被抛出： 

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)

而随着2.4的EclipseLink 

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，我尝试用以下QueryDSL语法来解决此问题： 

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

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

前半部分没有工作就像一个魅力，但后者没有返回预期的结果。 与目前在标签许可证withoutTags不排除在结果集，因为他们应该。

JPQL和SQL后者声明的样子： 

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))

该JPQL对我来说很好，但是如果一个许可证已与其相关联的多个标签的SQL明显失败。 所以我觉得这实际上是EclipseLink的翻译失败的情况下。 我要看看如果这是我使用的版本，一个已知的bug。 本文在某种程度上被支持JPQL使用查询标准API“不是会员”虽然只有同时使用creteria API，而不是JPQL出现的问题是如此。 这一错误翻译的EclipseLink仍然存在2.4 RC 2，这里是最后那做什么它意味着对于部分“无标签”解决方法： 

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，Tilmann

Answer 1:

你可以使用一个连接，而不是一个子选择？ 这将是更有效的为好。

您也可以尝试的EclipseLink 2.4，这可能没有这些问题。

QueryDSL支持联接，两者innerJoin（）和leftJoin（），你应该使用这个。



Answer 2:

目前，在的EclipseLink和QueryDSL臭虫否认直截了当语句用法： 

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

而是一种解决方法必须使用： 

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());


文章来源: QueryDSL duplicate identification variable/equality syntax error with any on Set?
标签： java jpa eclipselink querydsl
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