In the same table, use SQL to query the difference between today and yesterday, and then sort it. How to do it?

What is the faster way to write?

I am now using left join on the table itself, but the result seems wrong

SELECT

<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>

Reply content:

What is the faster way to write?

I am now using left join on the table itself, but the result seems wrong

SELECT

<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>

Suppose there is the following data tabletbl

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

Run

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;

Results

uid	diff
3	7
1	3
2	-2