问题
According to the tables:
USERS (user_name, email, balance)
How can I create a query that return the second highest user balance in the most efficient way ?
I successes to get this record (but not by the efficient way) with the query:
SELECT
*
FROM
(SELECT
us.*,
ROWNUM row_num
FROM
(SELECT
u.*
FROM
users u
ORDER BY
u.BALANCE DESC) us
WHERE
ROWNUM < 3)
WHERE
row_num > 1;
回答1:
I would use a window function:
select *
from (
select u.*, dense_rank() over (order by balance desc) as rnk
from users u
) t
where rnk = 2;
I don't think there will be a big performance difference to your query (especially not with an index on balance) but in my opinion it's easier to read and maintain.
回答2:
Try this:
SELECT *
FROM (SELECT *
FROM USERS
ORDER BY balance DESC
FETCH FIRST 2 ROWS ONLY
)
ORDER BY balance DESC
FETCH FIRST 1 ROWS ONLY
回答3:
This should work even in case of more than one user having same 2nd largest balance..
select * from USERS where balance IN
(select max(balance)
from (select balance from USERS
where balance NOT IN (select max(balance) from USERS))
);
来源：https://stackoverflow.com/questions/39100008/second-highest-value-from-oracle-dbs-table