How to order by the same filed twice by condition - sql

Question

I have this model :
public class Post {
private Long expiryDate; // A timestamp
}
And repository as :
#Repository
public interface PostRepository extends JpaRepository<Post, Long> {
Page<Post> findAllByOrderByExpiryDateDesc(Pageable pageable);
}
What I want to do:
When the expiry date has passed (post expired, current date greater than expiry date) => orderBy desc
Else, When the expiry date not yet passed (post not expired) => orderBy asc
Example: if I have the expiry date list: Monday, Tuesday, Wednesday, Thursday and Friday
and that today is Wednesday (Wednesday not yet expired).
Wanted result:
Wednesday
Thursday
Friday
Tuesday
Monday
Someone got any solution please?

Answer 1

You could try the following query:
SELECT p FROM Post p
ORDER BY CASE
WHEN p.expiryDate > CURRENT_TIMESTAMP THEN p.expiryDate
ELSE -p.expiryDate
END ASC
(you might need some casting to make it work, but since you didn't mention the RDBMS you're using, you'll need to check it out yourself)

Answer 2

I found the solution (with the help of #crizzis),
This Query work perfectly :
#Query(value = "from Post as p ORDER BY " +
"CASE WHEN p. expiryDate > :currentTime THEN -p.expiryDate ELSE null END desc," +
"CASE WHEN p. expiryDate < :currentTime THEN -p.expiryDate ELSE null END asc")
Page<Post> findAll(Long currentTime, Pageable pageable);
For native query (h2 database example)
SELECT * FROM Post p
ORDER BY
CASE WHEN p.expiryDate > currentTime THEN -p.expiryDate END desc,
CASE WHEN p.expiryDate < currentTime THEN -p.expiryDate END asc;