Parameter in like clause JPQL

问题

I am trying to write a JPQL query with a like clause:

LIKE \'%:code%\'

I would like to have code=4 and find

I cannot pass :code = \'%value%\'

namedQuery.setParameter(\"%\" + this.value + \"%\");

because in another place I need :value not wrapped by the % chars. Any help?

回答1:

If you do

LIKE :code

and then do

namedQuery.setParameter("code", "%" + this.value + "%");

Then value remains free from the '%' sign. If you need to use it somewhere else in the same query simply use another parameter name other than 'code' .

回答2:

I don't use named parameters for all queries. For example it is unusual to use named parameters in JpaRepository.

To workaround I use JPQL CONCAT function (this code emulate start with):

@Repository
public interface BranchRepository extends JpaRepository<Branch, String> {
    private static final String QUERY = "select b from Branch b"
       + " left join b.filial f"
       + " where f.id = ?1 and b.id like CONCAT(?2, '%')";
    @Query(QUERY)
    List<Branch> findByFilialAndBranchLike(String filialId, String branchCode);
}

I found this technique in excellent docs: http://openjpa.apache.org/builds/1.0.1/apache-openjpa-1.0.1/docs/manual/jpa_overview_query.html

回答3:

You could use the JPA LOCATE function.

LOCATE(searchString, candidateString [, startIndex]): Returns the first index of searchString in candidateString. Positions are 1-based. If the string is not found, returns 0.

FYI: The documentation on my top google hit had the parameters reversed.

SELECT 
  e
FROM 
  entity e
WHERE
  (0 < LOCATE(:searchStr, e.property))

回答4:

I don't know if I am late or out of scope but in my opinion I could do it like:

String orgName = "anyParamValue";

Query q = em.createQuery("Select O from Organization O where O.orgName LIKE '%:orgName%'");

q.setParameter("orgName", orgName);

回答5:

There is nice like() method in JPA criteria API. Try to use that, hope it will help.

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery criteriaQuery = cb.createQuery(Employees.class);
Root<Employees> rootOfQuery = criteriaQuery.from(Employees.class);
criteriaQuery.select(rootOfQuery).where(cb.like(rootOfQuery.get("firstName"), "H%"));

回答6:

Just leave out the ''

LIKE %:code%

回答7:

Use below JPQL query.

select i from Instructor i where i.address LIKE CONCAT('%',:address ,'%')");

Use below Criteria code for the same:

@Test public void findAllHavingAddressLike() {CriteriaBuilder cb = criteriaUtils.criteriaBuilder(); CriteriaQuery cq = cb.createQuery(Instructor.class); Root root = cq.from(Instructor.class); printResultList(cq.select(root) .where (cb.like(root.get(Instructor_.address), "%#1074%"))); }

来源：https://stackoverflow.com/questions/1341104/parameter-in-like-clause-jpql

标签

java

jpa

eclipselink

jpql

sql-like