JPQL Like Case Insensitive

问题

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

@Repository
public interface UserRepository extends JpaRepository<User, Long> {

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

}

I got an error: unexpected token: %. Where should I place '%'?

回答1:

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)

回答2:

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.

回答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");

来源：https://stackoverflow.com/questions/37178520/jpql-like-case-insensitive