Write an SQL query to find all dates' id with higher temperature compared to its previous dates (yesterday).
Return the result table in any order.
The query result format is in the following example:
문제링크
SELECT today.ID
FROM weather AS today
LEFT JOIN weather AS yesterday ON today.recorddate = yesterday.recorddate + INTERVAL 1 DAY
-- 이 문제는 id기준으로 SELF JOIN을 해주는 것이 아니라 recorddate기준으로 INTERVAL + 1을 활용해서 문제를 풀어주는 것임.
WHERE today.temperature > yesterday.temperature