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Eclipselink 2.5.2 giving exception on executing jpql having CASE expression [message #1434737]

Tue, 30 September 2014 14:04

Abhijeet Bongarde

Messages: 2
Registered: September 2014

Junior Member

I have a JPQL generated using eclipselink 2.5.2

SELECT t0 FROM DM_CONV_PUMP t0 WHERE (CASE WHEN t0.SUPPLIER_NAME LIKE 'bb%' THEN TRUE ELSE FALSE END) = true AND t0.TEIID_MULTI_DATA_SOURCE_COLUMN = 'EDM_sql'

I am getting following exception:
java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager:
Exception Description: Syntax error parsing [SELECT t0 FROM DM_CONV_PUMP t0 WHERE (CASE WHEN t0.SUPPLIER_NAME LIKE 'bb%' THEN TRUE ELSE FALSE END) = true AND t0.TEIID_MULTI_DATA_SOURCE_COLUMN = 'EDM_sql'].
[86, 86] The identifier THEN is missing from the WHEN clause.
[76, 80] The identification variable 'THEN' cannot be a reserved word.
org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1605)

This works in Eclipselink2.4.1.
While debugging, I found that 'DefaultEclipseLinkJPQLGrammar' is not able to parse 'THEN' keyword.

Is there any changes in Eclipselink2.5.2 which is causing this issue?

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Re: Eclipselink 2.5.2 giving exception on executing jpql having CASE expression [message #1435137 is a reply to message #1434737]

Wed, 01 October 2014 04:11

Chris Delahunt

Messages: 1389
Registered: July 2009

Senior Member

Odd, the parser seems to pick up 'THEN" as an identification variable. Please file a bug, and try using "CASE WHEN (t0.SUPPLIER_NAME LIKE 'bb%') THEN.." to see if that makes any difference. If not, an alternative workaround would be to use criteria api.

Best Regards,
Chris

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