按 somecolumn 分组以获得总值并按 somecolumn 以相反的顺序对值求和

Posted 2023-03-31

【中文标题】按 somecolumn 分组以获得总值并按 somecolumn 以相反的顺序对值求和

【英文标题】：group by somecolumn to get total value and sum the value in reverse order by somecolumn

【发布时间】：2019-05-29 02:32:49

【问题描述】：

表格&DDL：

|id |gooditem|amt  |
|---|--------|-----|
|1  |apple   |1000 |
|2  |apple   |2000 |
|3  |apple   |-3000|
|4  |apple   |500  |
|5  |apple   |1500 |
|6  |pen     |1000 |
|7  |pen     |2000 |
|8  |pen     |-3000|
|9  |pen     |500  |
|10 |pen     |2500 |

CREATE TABLE T
    ([id] int, [gooditem] varchar(5), [amt] int)
;

INSERT INTO T
    ([id], [gooditem], [amt])
VALUES
    (1, 'apple', 1000),
    (2, 'apple', 2000),
    (3, 'apple', -3000),
    (4, 'apple', 500),
    (5, 'apple', 1500),
    (6, 'pen', 1000),
    (7, 'pen', 2000),
    (8, 'pen', -3000),
    (9, 'pen', 500),
    (10, 'pen', 2500)
;

Test Demo Link

逻辑

产品foo总量为2000，pen为3000 按ID逆序求和

预期结果

|id |gooditem|amt  |formula                      |total_amt|
|---|--------|-----|-----------------------------|---------|
|1  |foo     |1000 |2000-1500-500--3000-2000-1000|0        |
|2  |foo     |2000 |2000-1500-500--3000-2000     |1000     |
|3  |foo     |-3000|2000-1500-500--3000          |3000     |
|4  |foo     |500  |2000-1500-500                |0        |
|5  |foo     |1500 |2000-1500                    |500      |
|6  |pen     |1000 |3000-1500-500--3000-2000-1000|0        |
|7  |pen     |2000 |3000-1500-500--3000-2000     |1000     |
|8  |pen     |-3000|3000-1500-500--3000          |2000     |
|9  |pen     |500  |3000-1500-500                |1000        |
|10 |pen     |1500 |3000-1500                    |1500      |

我的尝试和问题

我尝试使用rownumber() 来订购 id 并使用 id 总和数据。 但是我在gooditem to get total value and sum the value in reverse order by somecolumn的组上有问题

版本

Microsoft SQL Server 2008 R2 (RTM) - 10.50.1600.1 (X64)

---

我对 sql server 2008 的回答

with cte as (
    select row_number() over (partition by [gooditem] order by id desc) rnk
       ,sum(T1.amt) over (partition by T1.gooditem)  total_amt,* 
    from T T1
)
,cte2 as (
    select T1.gooditem ,T1.id,T1.rnk,T1.amt ,T1.total_amt -T1.amt as total_amt  from cte T1
    where T1.rnk = 1
    union all
    select T1.gooditem ,T1.id,T1.rnk,T1.amt ,T2.total_amt - T1.amt as total_amt from cte T1
    inner join cte2 T2 on T1.[gooditem] = T2.[gooditem] and T1.rnk = T2.rnk+1
)
select T1.id,T1.gooditem,T1.amt ,T1.total_amt
from cte2 T1
order by id 

db<>fiddle

【参考方案1】：

你正在寻找一个累计和：

select id, gooditem, amt,
       (2000 - sum(amt) over (partition by gooditem order by id desc)) as total_amt
from t
order by id;

Here 是一个 dbfiddle。

编辑：

如果您需要总数，您也可以使用窗口函数来计算：

select id, gooditem, amt,
       (sum(amt) over (partition by gooditem) -
        sum(amt) over (partition by gooditem order by id desc)
       ) as total_amt
from t
order by id;

db<>fiddle 非常相似。

【讨论】：

谢谢，但是不能使用常量2000，这是我之前试过的问题点

@ITWeiHan 。 . .那只需要累积总和。