동일한 테이블에서 SQL을 사용하여 오늘과 어제의 차이를 쿼리한 후 정렬하는 방법은 무엇입니까?

글을 빨리 쓰는 방법은 무엇인가요

이제 테이블 자체에 Left Join을 사용하고 있는데 결과가 잘못된 것 같습니다

선택

<code>            table.id ,SUM(table.s-yestoday.a) as sum
            FROM table
            LEFT JOIN table yestoday
            ON yestoday.uid = wy_appdata.uid
            WHERE table.year = '.$year.' AND table.month = '.$month.' AND table.day = '.$day.' AND yestoday.year = '.$bre_data['year'] .' AND yestoday.month ='.$bre_data['month'] .' AND yestoday.day = '.$bre_data['day'] .'
            GROUP BY table.uid
            ORDER BY sum DESC</code>

답글 내용:

글을 빨리 쓰는 방법은 무엇인가요

이제 테이블 자체에 Left Join을 사용하고 있는데 결과가 잘못된 것 같습니다

선택

<code>            table.id ,SUM(table.s-yestoday.a) as sum
            FROM table
            LEFT JOIN table yestoday
            ON yestoday.uid = wy_appdata.uid
            WHERE table.year = '.$year.' AND table.month = '.$month.' AND table.day = '.$day.' AND yestoday.year = '.$bre_data['year'] .' AND yestoday.month ='.$bre_data['month'] .' AND yestoday.day = '.$bre_data['day'] .'
            GROUP BY table.uid
            ORDER BY sum DESC</code>

다음과 같은 데이터 테이블이 있다고 가정해보자tbl

uid	s	date
1	5	2016-08-31
2	3	2016-08-31
3	7	2016-08-31
1	2	2016-08-30
2	5	2016-08-30
4	4	2016-08-30

달려

SELECT
  today.uid,
  today.s - IFNULL(yesterday.s, 0) AS diff
FROM
  (SELECT uid, SUM(s) AS s FROM tbl WHERE date='2016-08-31' GROUP BY uid) AS today
LEFT OUTER JOIN
  (SELECT uid, SUM(s) AS s FROM tbl WHERE date='2016-08-30' GROUP BY uid) AS yesterday
USING (uid)
ORDER BY diff DESC;

결과

uid	diff
3	7
1	3
2	-2