计算 table 中的数字变化

Question

我有一个 table 设置如下，它只捕获当天的总余额：

+--------+------------+--------+
|   id   |    date    | wallet |
+----+---------+----------+-----
| 10056  | 2022-03-01 |  120   | 
| 10056  | 2022-03-02 |   40   |
| 10056  | 2022-03-04 |   80   |
| 10056  | 2022-03-08 |   23   |
| 13572  | 2022-03-01 |   20   |
| 13572  | 2022-03-03 |   17   |
+--------+------------+--------+

请问是否可以只计算支出？我只能想着用'CASE'来停止计算比昨天小的数，但是我不知道如何逐行计算负数。这应该是结果：

-- March Spending per user
+---------+---------+
|   id    |  spend  |
+---------+---------+
| 10056   | 103     |
| 13572   | 3       |
+---------+---------+

我不确定我的SQL概念是否正确。

Answer 1

一种选择是计算连续钱包值之间的差异，然后排除负值（因为它们是钱包收入而不是支出）并求和。检查以下查询：

WITH cte AS (
    SELECT *, 
           wallet - LEAD(wallet) OVER(PARTITION BY id 
                                      ORDER     BY date)  AS diff
    FROM balances
)
SELECT id,
       SUM(diff)
FROM cte
WHERE diff > 0
GROUP BY id

试试看 here.

Answer 2

如果我能很好地解释你的问题，你必须分两个阶段解决问题：

将前一行的钱包值添加到每一行，以便您可以应用正确的逻辑。
应用允许您获取字段值并将它们分组的逻辑。

假设生成名为 prova 的原始 table，我的查询建议可以是：

select id, sum(case when prev_wallet>wallet then prev_wallet- wallet else 0 end) expense
from
(
select p.id, p.date, p.wallet, (select wallet from prova where id=p.id and date = (select max(date) date from prova where date<p.date and id=p.id)) prev_wallet
from prova p
order by p.id, p.date
) A
group by id

再见

计算 table 中的数字变化

Calculate the number change in the table

mysql

sql