JPQL Like Case Insensitive

Question

I want to search data in User table by name case insensitive.

@Repository
public interface UserRepository extends JpaRepository {

  @Query         


        

Answer 1

I am using Spring Boot 2.1.6, You can define query methods using Containing, Contains, and IsContaining as below:

List<User> findByNameContaining(String name);
List<User> findByNameContains(String name);
List<User> findByNameIsContaining(String name);

Case Insensitivity:

List<User> findByNameContainingIgnoreCase(String name);

OR you can also define as below as well:

@Query("select u from User u where lower(u.name) like lower(concat('%', :name,'%'))")
public List<User> findByName(@Param("name") String name);

The @Param annotation is important here because we're using a named parameter.

Answer 2

You can use the concat operator:

@Query("select u from User u where lower(u.name) like lower(concat('%', ?1,'%'))")
public List<User> findByNameFree(String name);

or with a named parameter:

@Query("select u from User u where lower(u.name) like lower(concat('%', :nameToFind,'%'))")
public List<User> findByNameFree(@Param("nameToFind") String name);

(Tested with Spring Boot 1.4.3)

Answer 3

You can use wildcard matching.

for example, i want to search name like haha,

@Query("select u from User u where lower(u.name) like :u_name")
public List<User> findByNameFree(@Param("u_name") String name);
List<User> users = userDao.findByNameFree("%haha");

Answer 4

If that is only what you want and you are using Spring Data JPA you don't need to write a query.

List<User> findByNameContainingIgnoreCase(String name);

Else you need to wrap the name attribute with % before you pass it to the method (putting those directly in the query will simply not work). Or don't use a query but use a specification or the Criteria API to create the query.