1831. Maximum Transaction Each Day ¶ SQL 1 2 3 4 5 6 7 8 9 10 11 12 13 14WITH RankedTransactions AS ( SELECT *, RANK() OVER( PARTITION BY DATE(day) ORDER BY amount DESC ) AS `rank` FROM Transactions ) SELECT transaction_id FROM RankedTransactions WHERE `rank` = 1 ORDER BY 1;