计算 SQL 中年初至今数字的周期值

Question

我已经累积了每个时期的年初至今值。但是我需要每个产品组/周期的周期值。

您是否知道 SQL 服务器中的解决方案以简单的方式实现此目的？

这是我的示例数据：

Product Group | Period | Amount
-------------------------------
Group 1       |2018/01 | 500
Group 1       |2018/02 | 740
Group 1       |2018/03 | 900
Group 1       |2018/04 | 930

结果应该是这样的：

Product Group | Period | Amount
-------------------------------
Group 1       |2018/01 | 500
Group 1       |2018/02 | 240
Group 1       |2018/03 | 160
Group 1       |2018/04 | 30

感谢您的帮助！菲利普

Answer 1

您可以像这样使用 LAG 函数：

SELECT [Product Group], Period,  
       Amount - LAG(Amount, 1,0) OVER (ORDER BY Period) AS Amount
FROM myTable

Answer 2

希望对您有所帮助：

Select 
ProductGroup, 
Period, 
Amount = Amount-ISNULL((Select Amount from TempTable where Period = (select max(Period) from TempTable where Period < Main.Period)),0)
From TempTable Main
Order by Period

我假设 Period 是日期时间类型 (2018-01-01)

Answer 3

查询中的一个小改动

SELECT [ProductGroup], Period,  
      Amount- LAG(Amount, 1,0) OVER (ORDER BY Period)   AS Amount
FROM tablename

Answer 4

希望对你有所帮助

WITH CTE AS (
SELECT
rownum = ROW_NUMBER() OVER (ORDER BY t.[Product Group],t.[Period]),
t.*
FROM YourTbl AS t
)

SELECT CTE.[Product Group],CTE.Period,-1*(ISNULL(prev.Amount,0)-CTE.Amount) AS Amount 
FROM CTE
LEFT JOIN CTE prev ON prev.rownum = CTE.rownum - 1

Answer 5

假设您不能使用 LAG（例如 MSSQL 2008）并且每个 ProductGroup 和 Period 可以有多个记录，您可以使用以下查询。

示例 table 和值：

CREATE TABLE GR (Product_Group VARCHAR(10), Period VARCHAR(6), Amount INT);

INSERT INTO GR VALUES ('Group 1',  '201801', 500)
, ('Group 1',  '201802', 740)
,('Group 1',  '201803', 900)
,('Group 1',  '201804', 930)
;
INSERT INTO GR VALUES ('Group 2',  '201801', 500)
, ('Group 2',  '201803', 800)
,('Group 2',  '201803', 1000)
,('Group 2',  '201804', 1200)
;

查询使用 GROUP BY（和 CTE 以简化阅读）对 ProductGroup 和 Period 进行分组，并使用 RowNumber 来查找以前的值（如果你有固定的时间段，即如果你想在月份值时显示每个月的记录缺少您可以使用理货日期 table)

WITH X AS (SELECT  Product_Group, Period, SUM(Amount) AS Amount_TOT
            , ROW_NUMBER() OVER (PARTITION BY Product_Group ORDER BY PERIOD) AS RN
            FROM GR GROUP BY Product_group, Period)  
SELECT Product_Group, Period, Amount_TOT,  Amount_TOT_PREC, Amount_TOT-ISNULL(Amount_TOT_PREC,0) AS Delta 
FROM (SELECT  A.Product_Group, A.Period, A.Amount_TOT, B.Amount_TOT AS Amount_TOT_PREC       
        FROM X A
        LEFT JOIN X B ON A.Product_Group=B.Product_Group AND A.RN-1 = B.RN
        ) C

输出

+---------------+--------+------------+-----------------+-------+
| Product_Group | Period | Amount_TOT | Amount_TOT_PREC | Delta |
+---------------+--------+------------+-----------------+-------+
| Group 1       | 201801 |        500 | NULL            |   500 |
| Group 1       | 201802 |        740 | 500             |   240 |
| Group 1       | 201803 |        900 | 740             |   160 |
| Group 1       | 201804 |        930 | 900             |    30 |
| Group 2       | 201801 |        500 | NULL            |   500 |
| Group 2       | 201803 |       1800 | 500             |  1300 |
| Group 2       | 201804 |       1200 | 1800            |  -600 |
+---------------+--------+------------+-----------------+-------+

使用 LAG（此函数 return 先前记录的值，请查看 Microsoft 文档）更快且更易读：

WITH X AS (SELECT Product_Group, Period
                  , SUM(Amount) AS Amount_TOT           
            FROM GR GROUP BY Product_group, Period)  
SELECT Product_Group, Period, Amount_TOT, AMOUNT_PREC
       , Amount_TOT-ISNULL(AMOUNT_PREC,0) AS Delta 
FROM (SELECT  Product_Group, Period, Amount_TOT 
              , LAG(Amount_TOT) OVER (PARTITION BY Product_Group 
                 ORDER BY PERIOD) AS AMOUNT_PREC     
        FROM X) A;

与上面相同的输出

Answer 6

我会使用 lag() 函数：

select [Product Group], Period,  
       Amount - LAG(Amount, 1,0) OVER (ORDER BY Period) AS Amount
from table t'

如果你有旧版本的 SQL，那么你也可以使用 apply :

select t.product, t.period, coalesce(t.amount-t1.amount, t.amount) as amount
from table t outer apply
     ( select top (1) t1.*
       from table t1
       where t1.product = t.product and 
             t1.period < t.period
             order by t1.period desc
     ) t1;

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