功能和操作员无法在Eclipselink中工作

Question

我有这个问题,REPLACE功能不适用于eclipselink版本2.5.2.

这是我的代码:

String sSql = " SELECT e FROM br.com.megasoft.protocolo.entity.Assunto e  WHERE  ( REPLACE(REPLACE(REPLACE( UPPER(e.titulo), '/', ''), '-', ''), '.', '') LIKE  UPPER('A') )";
TypedQuery<?> query = getEntityManager().createQuery(sSql, Class.forName(this.tabela));

第二个参数的值为: class br.com.megasoft.protocolo.entity.Assunto

它给出了这个例外:

Caused by: Exception [EclipseLink-0] (Eclipse Persistence Services - 2.5.2.v20140319-9ad6abd): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Syntax error parsing [ FROM br.com.megasoft.protocolo.entity.Assunto e  WHERE  ( REPLACE(REPLACE(REPLACE( UPPER(e.titulo), '/', ''), '-', ''), '.', '') LIKE  UPPER( :valorPesq10) ) ]. 
[54, 150] The expression is not a valid conditional expression.
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.buildException(HermesParser.java:155)
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.validate(HermesParser.java:334)
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.populateQueryImp(HermesParser.java:278)
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.buildQuery(HermesParser.java:163)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:142)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:116)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:102)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:86)
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1603)
    ... 50 more

当我使用Hibernate时它完全正常.

当我运行简单的SQL,没有REPLACE,它工作正常.

Obs:Eclipselink 2.5.2正在使用JPA 2.1.我正在使用Tomcat8,Eclipse Kepler.

Answer 1

I have found the solution, thank you guys for the useful informations.

The sql needs to be this way:

SELECT e FROM br.com.megasoft.protocolo.entity.Assunto e  WHERE  FUNCTION('REPLACE',  e.titulo, '/', '') LIKE  UPPER('A')

This way, I will be using the FUNCTION defined in JPA 2.1 spec.