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Spring list of hibernation patterns as parameter

I am trying to execute this query: Code:

this.getHibernateTemplate()
      find("select distinct ci.customer " +
             "from CustomerInvoice ci " +
              "where ci.id in (?) " , ids);

with identifiers like List, id is of type Long

when executed i get exception

the 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
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5 answers

In addition to the mR_fr0g answer, this also works:

this.getHibernateTemplate() 
      findByNamedParam("select distinct ci.customer " + 
             "from CustomerInvoice ci " + 
              "where ci.id in (:ids) ", "ids", ids);
+7
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If you want to add a list to a sentence, it is best to use a named parameter. It is done like this.

Query q = this.getHibernateTemplate().getSession().createQuery("select distinct ci.customer " +
             "from CustomerInvoice ci " +
              "where ci.id in (:idsParam) ");
q.setParameter("idsParam", ids);
List<Customer> = q.getResultList();
+9
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API- Hibernate Criteria, "in".

:

Btw. , ! ( API)

+3

You can use the parameter list for inlcude in your query with "IN" and "setParameterList"

List<Long> ids= new ArrayList<Long>();

Query query = getSession().createQuery("select distinct ci.customer from CustomerInvoice ci where ci.id in (:ids) ");
query.setParameterList("ids", ids);
query.executeUpdate();
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   ProjectionList projList =
   Projections.projectionList().add("customer","customer");  
   List<Long> ids = ids;
   Criteria criteria = hibernateTemplate.getSessionFactory().getCurrentSession()
      .createCriteria(CustomerInvoice.class)
      .add(Restrictions.in("id",ids))
      .setProjection(projList); List<Long> listOfIds = criteria.list();
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