EJB QL "LIKE" PROBLEM

Question

Hai All.
I got an error when put this EJB-QL syntax :
<query>
<query-method>
<method-name>findByName</method-name>
<method-params>
<method-param>java.lang.String</method-param>
</method-params>
</query-method>
<ejb-ql>SELECT DISTINCT object (a) FROM DummyData a WHERE a.name LIKE ?1</ejb-ql>
</query>
Seems like this method cannot take an arguments, because it runs well when I change the <ejb-ql> syntax with harcode query like this :
<ejb-ql>SELECT DISTINCT object (a) FROM DummyData a WHERE a.name LIKE %FOO%</ejb-ql>
Is there something wrong with my first syntax?
or the EJB-QL cannot take an arguments for LIKE operator in EJB 2.0??
Below is the error output
Auto-deploying file:/C:/NS/jdev/TAX_WORKSPACE/TAX_PROJECT/classes/ (No previous deployment found)... Encountered "a . birthdate LIKE ?1" at line 1, column 7.
EJB QL method : public abstract java.util.Collection gov.tax.business.registration.IEbDummyLocalHome.findByName(java.lang.String) throws javax.ejb.FinderException
com.sun.ejb.ejbql.EjbQLQuery
com.sun.ejb.ejbql.parser.EjbQLParser.parse(com.sun.ejb.ejbql.SymbolTable, com.sun.enterprise.deployment.PersistenceDescriptor, java.lang.reflect.Method, java.lang.String)
Thanx in advance

Answer 1

Hi,
You can achieve this by overriding EJB-QL in orion-ejb-jar.xml file. Please look at
http://otn.oracle.com/docs/products/ias/doc_library/903doc_otn/generic.903/a97677/ql.htm#1005714
So you can have something like
<finder-method query="$name like $1"> in orion-ejb-jar.xml
and a dummy (but valid) ejb-ql in ejb-jar.xml, like
<ejb-ql>SELECT DISTINCT object (a) FROM DummyData a WHERE a.name =?1</ejb-ql>
Hope this helps,
Neelesh
OTN Team @ IDC