SQL LAG 函数和分组/排序

Question

考虑下表：

    +--------------------------------------------------+
    |                        Users                     |                  
    +-------+---------+----------+----------+----------+
    |acc_num| user_id |   date   |  amount  |    sum   |
    +-------+---------+----------+----------+----------+
    |   a1  |    u1   | 20201209 |    20    |   null   |
    |   a1  |    u1   | 20201209 |    20    |          |
    |   a1  |    u1   | 20201209 |    20    |          |
    +-------+---------+----------+----------+----------+
    |   a1  |    u2   | 20201208 |    30    |    50    |        Correct
    |   a1  |    u2   | 20201208 |    30    |          |
    |   a1  |    u2   | 20201208 |    30    |          |
    +-------+---------+----------+----------+----------+

如上所示，我想根据今天和前一天计算总和。因为我在 Postgres 上，所以我正在使用 LAG 功能。但是，即使我基于 acc_num、user_id 和 date 进行分组和排序，但是当添加新的 sum 列时，值有时是 60，有时是 50。所以有时我得到上述结果，有时我得到以下结果：

    +--------------------------------------------------+
    |                        Users                     |                  
    +-------+---------+----------+----------+----------+
    |acc_num| user_id |   date   |  amount  |    sum   |
    +-------+---------+----------+----------+----------+
    |   a1  |    u1   | 20201209 |    20    |   null   |
    |   a1  |    u1   | 20201209 |    20    |          |
    |   a1  |    u1   | 20201209 |    20    |          |
    +-------+---------+----------+----------+----------+
    |   a1  |    u2   | 20201208 |    30    |          |
    |   a1  |    u2   | 20201208 |    30    |    60    |        Wrong
    |   a1  |    u2   | 20201208 |    30    |          |
    +-------+---------+----------+----------+----------+

或者我明白了：

    +--------------------------------------------------+
    |                        Users                     |                  
    +-------+---------+----------+----------+----------+
    |acc_num| user_id |   date   |  amount  |    sum   |
    +-------+---------+----------+----------+----------+
    |   a1  |    u1   | 20201209 |    20    |   null   |
    |   a1  |    u1   | 20201209 |    20    |          |
    |   a1  |    u1   | 20201209 |    20    |          |
    +-------+---------+----------+----------+----------+
    |   a1  |    u2   | 20201208 |    30    |          |
    |   a1  |    u2   | 20201208 |    30    |          |
    |   a1  |    u2   | 20201208 |    30    |    60    |        Wrong
    +-------+---------+----------+----------+----------+

现在我的 SQL 是这样的：

    SELECT acc_num, user_id, date
      (CASE 
        WHEN (amount > 0) 
        THEN LAG(amount, 1) OVER (ORDER BY acc_num, user_id, date) + amount 
        ELSE NULL 
        END
      ) AS sum
    FROM Users
    GROUP BY acc_num, user_id, date
    ORDER BY acc_num, user_id, date

我也试过了：

    SELECT acc_num, user_id, date
      (CASE 
        WHEN (amount > 0) 
        THEN LAG(amount, 1) OVER (ORDER BY date) + amount 
        ELSE NULL 
        END
      ) AS sum
    FROM Users
    GROUP BY acc_num, user_id, date
    ORDER BY acc_num, user_id, date

有什么想法吗？谢谢。

Answer 1

使用lead 尝试以下操作，这里是demo。

select
    acc_num, 
    user_id, 
    date, 
    amount,
    sum(case when (nm is not null and amount <> nm) then nm + amount end) over (partition by acc_num, user_id) as sum 
from
(
    select
        *,
        lead(amount) over (order by date) as nm
    from myTable
) t

输出：

| acc_num | user_id | date     | amount | sum |
| ------- | ------- | -------- | ------ | --- |
| a1      | u1      | 20201209 | 20     |     |
| a1      | u1      | 20201209 | 20     |     |
| a1      | u1      | 20201209 | 20     |     |
| a1      | u2      | 20201208 | 30     | 50  |
| a1      | u2      | 20201208 | 30     | 50  |
| a1      | u2      | 20201208 | 30     | 50  |

Answer 2

如果我理解正确，您只需要前 天 的一个值。出于几个原因，这有点棘手，但非常重要的是，因为您没有稳定的行编号 - 也就是说，存在重复。

我建议：

select t.*,
       (case when lag(date) over (order by seqnum) < date
             then amount + lag(amount) over (order by seqnum)
        end)
from (select t.*, row_number() over (order by date, acc_num, user_id) as seqnum
      from mytable t
     ) t
order by seqnum;

Here 是一个 dbfiddle。

如果您希望在所有三行上都使用此功能，您也可以使用 last_value() 和 range 窗口框架：

select t.*,
       amount + last_value(amount) over (order by date range between unbounded preceding and interval '1 day' preceding)
from mytable t