使用 mysql 更新減去上一行

Question

數據庫中的表

日期	數量	姓名	ID
12-01	10.00	約翰·多伊	1002
12-02	10.00	約翰·多伊	1003
11-01	50.00	約翰·多伊	8976
11-02	50.00	約翰·多伊	8977
09-01	50.00	約翰·多伊	6788
09-02	50.00	約翰·多伊	6799
09-02	50.00	艾麗西婭·多伊	6800

結果應該是這樣的：

月	數量	總增加/減少	姓名
九月	100.00	100.00	約翰·多伊
十一月	100.00	0	約翰·多伊
十二月	20.00	-80.00	約翰·多伊

Answer 1

你可以用這樣的查詢來做到這一點

SELECT a1.pdate
 , SUM(a1.amount) AS amount
 , SUM(a1.total) AS total
 , a1.username AS username
FROM (
  SELECT a.*, COALESCE(a.amount+LAG(amount) OVER (ORDER BY pdate),0) AS total
  FROM am a 
  ORDER BY pdate
) AS a1
GROUP BY username,MONTH(pdate);

樣本

MariaDB [bernd]> SELECT * from am;
+----+------------+--------+----------+
| id | pdate      | amount | username |
+----+------------+--------+----------+
|  1 | 2021-12-01 |  10.00 | john doe |
|  2 | 2021-12-02 |  10.00 | john doe |
|  3 | 2021-11-01 |  50.00 | john doe |
|  4 | 2021-11-02 |  50.00 | john doe |
+----+------------+--------+----------+
4 rows in set (0.02 sec)

MariaDB [bernd]> SELECT a1.pdate
    ->  , SUM(a1.amount) AS amount
    ->  , SUM(a1.total) AS total
    ->  , a1.username AS username
    -> FROM (
    ->   SELECT a.*, COALESCE(a.amount+LAG(amount) OVER (ORDER BY pdate),0) AS total
    ->   FROM am a 
    ->   ORDER BY pdate
    -> ) AS a1
    -> GROUP BY username,MONTH(pdate);
+------------+--------+--------+----------+
| pdate      | amount | total  | username |
+------------+--------+--------+----------+
| 2021-11-01 | 100.00 | 100.00 | john doe |
| 2021-12-01 |  20.00 |  80.00 | john doe |
+------------+--------+--------+----------+
2 rows in set (0.00 sec)

MariaDB [bernd]> 

Answer 2

看起來這就是你要找的東西：

SELECT MONTHNAME(Last_trans) AS Month,
       name,
       total,
       prevTotal,
       total-IFNULL(prevTotal,0) AS Diff
FROM
(SELECT MAX(date) Last_trans,
       MONTH(date) mth,
       name,
       SUM(amount) total,
       LAG(SUM(amount)) OVER (PARTITION BY name ORDER BY MONTH(date)) prevTotal
FROM mytable
GROUP BY MONTH(date), name) V;

演示小提琴