同一个表，用sql查询今天和昨天的差值，然后排序，要怎么做

比较快的写法是什么

我现在用left join 这个表本身，但是结果好像不对

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>

回复内容：

比较快的写法是什么

我现在用left join 这个表本身，但是结果好像不对

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>

假设有如下数据表tbl

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

运行

<code class="sql">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;</code>

结果

uid	diff
3	7
1	3
2	-2