Spring hibernate模板列表作为参数 [英] Spring hibernate template list as a parameter
本文介绍了Spring hibernate模板列表作为参数的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
代码:
$ p $
this.getHibernateTemplate()
从CustomerInvoice ci+
找到(select distinct ci.customer+
where where ci.id in(?),ids);
以id作为List,id的类型为Long
执行时我得到异常
代码:
java.lang.ClassCastException:java.util.ArrayList不能转换为java.lang.Long
在org.hibernate.type.LongType.set(LongType.java:42)
在org.hibernate .type.NullableType.nullSafeSet(NullableType.java:136)
at org.hibernate.type.NullableType.nullSafeSet(NullableType.java:116)
at org.hibernate.param.PositionalParameterSpecification.bind(PositionalParameterSpecification .java:39)
at org.hibernate.loader.hql.QueryLoader.bindParameterValues(QueryLoader.java:491)
at org.hibernate.loader.Loader.prepareQueryStatement(Loader.java:1563)
在org.hibernate.loader.Loader.doQuery(Loader.java:673)
在org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:236)
在org.hibernate。 loader.Loader.doList(Loader.java:2220)
在org.hibernate.loader.Loader.listI gnoreQueryCache(Loader.java:2104)
在org.hibernate.loader.Loader.list(Loader.java:2099)
在org.hibernate.loader.hql.QueryLoader.list(QueryLoader.java:
at org.hibernate.hql.ast.QueryTranslatorImpl.list(QueryTranslatorImpl.java:338)
at org.hibernate.engine.query.HQLQueryPlan.performList(HQLQueryPlan.java:172)
在org.hibernate.impl.SessionImpl.list(SessionImpl.java:1121)
在org.hibernate.impl.QueryImpl.list(QueryImpl.java:79)
在org.springframework.orm .hibernate3.HibernateTemplate $ 29.doInHibernate(HibernateTemplate.java:849)
at org.springframework.orm.hibernate3.HibernateTemplate.execute(HibernateTemplate.java:372)
at org.springframework.orm.hibernate3。 HibernateTemplate.find(HibernateTemplate.java:840)
at org.springframework.orm.hibernate3.HibernateTemplate.find(HibernateTemplate.java:836)
at
解决方案
除了mR_fr0g的回答,这一个也适用:
this.getHibernateTemplate()
findByNamedParam(从CustomerInvoice中选择不同的ci.customer+
ci+
where ci.id in(:ids),ids,ids);
i'm trying to execute this query : Code:
this.getHibernateTemplate()
find("select distinct ci.customer " +
"from CustomerInvoice ci " +
"where ci.id in (?) " , ids);
with ids as a List, id is of type Long
when executing i get exception
Code:
java.lang.ClassCastException: java.util.ArrayList cannot be cast to java.lang.Long
at org.hibernate.type.LongType.set(LongType.java:42)
at org.hibernate.type.NullableType.nullSafeSet(NullableType.java:136)
at org.hibernate.type.NullableType.nullSafeSet(NullableType.java:116)
at org.hibernate.param.PositionalParameterSpecification.bind(PositionalParameterSpecification.java:39)
at org.hibernate.loader.hql.QueryLoader.bindParameterValues(QueryLoader.java:491)
at org.hibernate.loader.Loader.prepareQueryStatement(Loader.java:1563)
at org.hibernate.loader.Loader.doQuery(Loader.java:673)
at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:236)
at org.hibernate.loader.Loader.doList(Loader.java:2220)
at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2104)
at org.hibernate.loader.Loader.list(Loader.java:2099)
at org.hibernate.loader.hql.QueryLoader.list(QueryLoader.java:378)
at org.hibernate.hql.ast.QueryTranslatorImpl.list(QueryTranslatorImpl.java:338)
at org.hibernate.engine.query.HQLQueryPlan.performList(HQLQueryPlan.java:172)
at org.hibernate.impl.SessionImpl.list(SessionImpl.java:1121)
at org.hibernate.impl.QueryImpl.list(QueryImpl.java:79)
at org.springframework.orm.hibernate3.HibernateTemplate$29.doInHibernate(HibernateTemplate.java:849)
at org.springframework.orm.hibernate3.HibernateTemplate.execute(HibernateTemplate.java:372)
at org.springframework.orm.hibernate3.HibernateTemplate.find(HibernateTemplate.java:840)
at org.springframework.orm.hibernate3.HibernateTemplate.find(HibernateTemplate.java:836)
at
解决方案
In addition to mR_fr0g's answer, this one also works:
this.getHibernateTemplate()
findByNamedParam("select distinct ci.customer " +
"from CustomerInvoice ci " +
"where ci.id in (:ids) ", "ids", ids);
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