使用 MySQL 中的 ID 和日期进行累计和
Making cumulative sum with ids and dates in MySQL
假设我有这个：
id
transactiondate
transactiontag
clientid
transactionvalue
1
2020-12-28
pizza
0
29.99
2
2020-12-29
payment
0
-39.99
3
2020-12-28
cheese
0
-109.99
4
2020-12-28
cheese
0
-109.99
5
2020-12-28
pizza
1
-19.99
6
2020-12-28
cheese
1
-18.99
7
2020-12-28
salary
1
58.99
8
2020-12-29
salary
1
38.99
所以我有所有交易的 ID，交易的时间和支付的金额，我需要做的是每个日期和每个人的累计总和，所以答案应该是这样的：
这个代码 -> SELECT distinct accountid, transactiondate, SUM(transactionvalue) OVER(ORDER BY transactiondate) AS cumulative_sum FROM acc_transactions where accountid = 1; 给我这个
clientid
date
cumulative_sum
1
2020-12-28
20.01
1
2020-12-29
59.00
问题是，对于客户端 1 和 0，我都需要这个，到目前为止我已经尽了一切努力！现在需要英雄哈哈！
正确答案示例↓
clientid
date
cumulative_sum
1
2020-12-28
20.01
1
2020-12-29
59.00
0
2020-12-28
-189.99
0
2020-12-29
-229.98
首先你想要一个每日汇总：
select accountid, transactiondate,
sum(transactionvalue) as transactionvalue
from acc_transactions
group by accountid, transactiondate
order by accountid, transactiondate
然后，您可以在其上使用 window 函数：
select accountid, transactiondate,
sum(sum(transactionvalue))
over(partition by accountid order by transactiondate)
as cumulative_sum
from acc_transactions
group by accountid, transactiondate
order by accountid, transactiondate
假设我有这个：
| id | transactiondate | transactiontag | clientid | transactionvalue |
|---|---|---|---|---|
| 1 | 2020-12-28 | pizza | 0 | 29.99 |
| 2 | 2020-12-29 | payment | 0 | -39.99 |
| 3 | 2020-12-28 | cheese | 0 | -109.99 |
| 4 | 2020-12-28 | cheese | 0 | -109.99 |
| 5 | 2020-12-28 | pizza | 1 | -19.99 |
| 6 | 2020-12-28 | cheese | 1 | -18.99 |
| 7 | 2020-12-28 | salary | 1 | 58.99 |
| 8 | 2020-12-29 | salary | 1 | 38.99 |
所以我有所有交易的 ID，交易的时间和支付的金额，我需要做的是每个日期和每个人的累计总和，所以答案应该是这样的：
这个代码 -> SELECT distinct accountid, transactiondate, SUM(transactionvalue) OVER(ORDER BY transactiondate) AS cumulative_sum FROM acc_transactions where accountid = 1; 给我这个
| clientid | date | cumulative_sum |
|---|---|---|
| 1 | 2020-12-28 | 20.01 |
| 1 | 2020-12-29 | 59.00 |
问题是，对于客户端 1 和 0，我都需要这个，到目前为止我已经尽了一切努力！现在需要英雄哈哈！
正确答案示例↓
| clientid | date | cumulative_sum |
|---|---|---|
| 1 | 2020-12-28 | 20.01 |
| 1 | 2020-12-29 | 59.00 |
| 0 | 2020-12-28 | -189.99 |
| 0 | 2020-12-29 | -229.98 |
首先你想要一个每日汇总：
select accountid, transactiondate,
sum(transactionvalue) as transactionvalue
from acc_transactions
group by accountid, transactiondate
order by accountid, transactiondate
然后，您可以在其上使用 window 函数：
select accountid, transactiondate,
sum(sum(transactionvalue))
over(partition by accountid order by transactiondate)
as cumulative_sum
from acc_transactions
group by accountid, transactiondate
order by accountid, transactiondate