CODEWITHSUNDEEP
hibernate
EMM
EMM

Reputation: 1812

Hibernate: Parameter value did not match expected type?

When I try to run:

List<MappingInstruction> result = emFactory.get()
          .createQuery("FROM " + MappingInstruction.class.getSimpleName() + " c where c.mapping = :mapping",
              MappingInstruction.class).setParameter("mapping", mappingInfo.getId()).getResultList();

I get:

Service exception executing action "MappingAddAction", java.lang.IllegalArgumentException: Parameter value [5118] did not match expected type [com.vernuso.trust.server.domain.clientimport.MappingInfo (n/a)]

I debugged in Eclipse and mappingInfo.getId() does return 5118.

Can someone please help me understand why it expects the type MappingInfo instead of long?

I have two tables as shown in the image below. MappingInfo has One-To-Many relationship with MappingInstruction table.

enter image description here

Upvotes: 0

Views: 8535

Answers (1)

CrawlingKid
CrawlingKid

Reputation: 994

Look closely:

List<MappingInstruction> result = emFactory.get()
          .createQuery("FROM " + MappingInstruction.class.getSimpleName() + " c where c.mapping = :mapping",
              MappingInstruction.class).setParameter("mapping", mappingInfo.getId()).getResultList();

You are fetching from MappingInstruction Entity. In the where condition, c.mapping is of type MappingInfo. Please check the datatype of mapping variable inside MappingInstruction class.

Because c.mapping is of type MappingInfo,but you are passing mappingInfo.getId() which is long, the ava.lang.IllegalArgumentException throws.

So, Correction is as follows:

List<MappingInstruction> result = emFactory.get()
              .createQuery("FROM " + MappingInstruction.class.getSimpleName() + " c where c.mapping = :mapping",
                  MappingInstruction.class).setParameter("mapping", mappingInfo).getResultList();

Upvotes: 1

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