Spring data JPA query with parameter properties

SpringJpaSpring DataSpring Data-Jpa

Spring Problem Overview


What is the simplest way of declaring a Spring data JPA query that uses properties of an input parameter as query parameters?

For example, suppose I have an entity class:

public class Person {
    @Id
    private long id;

    @Column
    private String forename;

    @Column
    private String surname;
}

and another class:

public class Name {
    private String forename;
    private String surname;

    [constructor and getters]
}

... then I would like to write a Spring data repository as follows:

public interface PersonRepository extends CrudRepository<Person, Long> {
    @Query("select p from Person p where p.forename = ?1.forename and p.surname = ?1.surname")
    findByName(Name name);
}

... but Spring data / JPA doesn't like me specifying property names on the ?1 parameter.

What is the neatest alternative?

Spring Solutions


Solution 1 - Spring

This link will help you: Spring Data JPA M1 with SpEL expressions supported. The similar example would be:

@Query("select u from User u where u.firstname = :#{#customer.firstname}")
List<User> findUsersByCustomersFirstname(@Param("customer") Customer customer);

https://spring.io/blog/2014/07/15/spel-support-in-spring-data-jpa-query-definitions

Solution 2 - Spring

Define the query method with signatures as follows.

@Query(select p from Person p where p.forename = :forename and p.surname = :surname)
User findByForenameAndSurname(@Param("surname") String lastname,
                             @Param("forename") String firstname);
}

For further details, check the Spring Data JPA reference

Solution 3 - Spring

What you want is not possible. You have to create two parameters, and bind them separately:

select p from Person p where p.forename = :forename and p.surname = :surname
...
query.setParameter("forename", name.getForename());
query.setParameter("surname", name.getSurname());

Solution 4 - Spring

You could also solve it with an interface default method:

 @Query(select p from Person p where p.forename = :forename and p.surname = :surname)
User findByForenameAndSurname(@Param("surname") String lastname,
                         @Param("forename") String firstname);

default User findByName(Name name) {
  return findByForenameAndSurname(name.getLastname(), name.getFirstname());
}

Of course you'd still have the actual repository function publicly visible...

Solution 5 - Spring

You can try something like this:

public interface PersonRepository extends CrudRepository<Person, Long> {
       @Query("select p from Person AS p"
       + " ,Name AS n"  
       + " where p.forename = n.forename "
       + " and p.surname = n.surname"
       + " and n = :name")
       Set<Person>findByName(@Param("name") Name name);
    }

Solution 6 - Spring

if we are using JpaRepository then it will internally created the queries.

Sample

> findByLastnameAndFirstname(String lastname,String firstname) > > findByLastnameOrFirstname(String lastname,String firstname) > > findByStartDateBetween(Date date1,Date2) > > findById(int id)

Note

if suppose we need complex queries then we need to write manual queries like

@Query("SELECT salesOrder FROM SalesOrder salesOrder WHERE salesOrder.clientId=:clientId AND salesOrder.driver_username=:driver_username AND salesOrder.date>=:fdate AND salesOrder.date<=:tdate ")
 @Transactional(readOnly=true)
 List<SalesOrder> findAllSalesByDriver(@Param("clientId")Integer clientId, @Param("driver_username")String driver_username, @Param("fdate") Date fDate, @Param("tdate") Date tdate);

Solution 7 - Spring

Are you working with a @Service too? Because if you are, then you can @Autowired your PersonRepository to the @Service and then in the service just invoke the Name class and use the form that @CuriosMind... proposed:

@Query(select p from Person p where p.forename = :forename and p.surname = :surname)
User findByForenameAndSurname(@Param("surname") String lastname,
                         @Param("forename") String firstname);
}

and when invoking the method from the repository in the service, you can then pass those parameters.

Solution 8 - Spring

The simplicity of Spring Data JPA is that it tries to interpret from the name of the function in repository without specifying any additional @Query or @Param annotations. If you are supplying the complete name, try to break it down as firstname and lastname and then use something like this -

HotelEntity findByName(String name);

My HotelEntity does contain the field name so JPA tries to interpret on its own to infer the name of the field I am trying to query on and create a subsequent query internally. Some more evidence from JPA documentation - enter image description here

Further details - here

Solution 9 - Spring

    for using this, you can create a Repository for example this one:
    Member findByEmail(String email);

    List<Member> findByDate(Date date);
    // custom query example and return a member
   @Query("select m from Member m where m.username = :username and m.password=:password")
        Member findByUsernameAndPassword(@Param("username") String username, @Param("password") String password);

Solution 10 - Spring

@Autowired
private EntityManager entityManager;

@RequestMapping("/authors/{fname}/{lname}")
    public List actionAutherMulti(@PathVariable("fname") String fname, @PathVariable("lname") String lname) {
        return entityManager.createQuery("select A from Auther A WHERE A.firstName = ?1 AND A.lastName=?2")
                .setParameter(1, fname)
                .setParameter(2, lname)
                .getResultList();
    }

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