Like 子句中的参数

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' .

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

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))

Just leave out the ''

LIKE %:code%

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);

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%"));

1. Use below JPQL query.

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

2. Use below Criteria code for the same:

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

Use JpaRepository or CrudRepository as repository interface:

@Repository
public interface CustomerRepository extends JpaRepository<Customer, Integer> {


@Query("SELECT t from Customer t where LOWER(t.name) LIKE %:name%")
public List<Customer> findByName(@Param("name") String name);


}




@Service(value="customerService")
public class CustomerServiceImpl implements CustomerService {


private CustomerRepository customerRepository;
    

//...


@Override
public List<Customer> pattern(String text) throws Exception {
return customerRepository.findByName(text.toLowerCase());
}
}