如何创建 table + 计算成本涉及 2 列

Question

我想按阶段计算成本，但它不起作用我的数据如下。我正在使用此命令来计算成本

SELECT Daily.dayno, activity.class, activity.phase,
    activity.duration/sum(activity.duration)*daily.daycost as cost
from Daily INNER JOIN
(
    Select activity.dailyuid, sum(activity.duration) duration
    group by dailyUID
) Activity
on activity.dailyuid=daily.dailyuid
group by class, phase

显然上述查询不起作用。我想要的是按阶段计算成本，例如

Dailyuid DD1 中的阶段 SS1 = 4/12*1000，其中 4 是 SS1 的总小时数，12 是 dailyuid DD1 的总小时数，1000 是 dailyuid DD1 的成本。

另一个例子是相位ST2 = 8/12*1000 IH=10.6/11.6*2000 等等

每天Table

DailyUID | Dayno | Daycost |
---------|-------|---------|
DD1      | 1     | 1000    | 
DD2      | 2     | 2000    |
DD3      | 3     | 3000    |

Activity Table

DailyUID | CLASS | PHASE | Duration |
-----------------------------------
DD1      | TS    | SS1   | 3        |
DD1      | TS    | SS1   | 1        |
DD1      | TS    | ST2   | 2        |
DD1      | P     | ST2   | 6        |
DD2      | P     | IH1   | 6.6      |
DD2      | U     | IH1   | 4        |
DD2      | TS    | IH2   | 1        |
DD3      | TU    | SC1   | 7        |
DD3      | P     | SC2   | 8        |
DD3      | U     | CMPLT | 3        |

Answer 1

如果我没理解错的话，你需要在加入之前pre-aggregate表格。需要一个额外的聚合：

select ap.phase,
       (ap.duration / ad.duration) * d.daycost as cost
from (SELECT d.dailyUid, sum(daycost) as daycost
      FROM Daily d
      GROUP BY d.dailyuid
     ) d INNER JOIN
     (Select a.dailyuid, sum(a.duration) as duration
      from activity a
      group by dailyUID
     ) ad
     on ad.dailyuid = d.dailyuid join
     (select a.dailyuid, a.phase, sum(a.duration) as duration
      from activity a
      group by a.dailyuid, a.phase
     ) ap
     on ad.dailyuid = ap.dailyuid
group by phase;

SQL Fiddle 是 here.

Answer 2

我想我用 SUM() OVER (...):

做到了

设置：

-- drop table Daily
create table Daily
(
   DailyUID VARCHAR(6),
   DayNo INT,
   DayCost NUMERIC(18, 2)
)
go

insert into Daily values ('DD1', 1, 1000),
    ('DD2', 2     , 2000),
    ('DD1', 3     , 3000)
go

-- drop table ActivityTable
create table ActivityTable
(
    DailyUID VARCHAR(6),
    Class VARCHAR(2),
    Phase VARCHAR(6),
    Duration NUMERIC(5, 1)
)
go

insert into ActivityTable values 
('DD1'   , 'TS'    , 'SS1'   , 3        ),
('DD1'   , 'TS'    , 'SS1'   , 1        ),
('DD1'   , 'TS'    , 'ST2'   , 2        ),
('DD1'   , 'P'     , 'ST2'   , 6        ),
('DD2'   , 'P'     , 'IH1'   , 6.6      ),
('DD2'   , 'U'     , 'IH1'   , 4        ),
('DD2'   , 'TS'    , 'IH2'   , 1        ),
('DD3'   , 'TU'    , 'SC1'   , 7        ),
('DD3'   , 'P'     , 'SC2'   , 8        ),
('DD3'   , 'U'     , 'CMPLT' , 3        )
GO

脚本：（仅适用于 SQL 服务器，请参阅下面的 MySql 版本）

    ;WITH Sums AS (
        SELECT DISTINCT AT.DailyUID, AT.Phase, 
            SUM(AT.Duration) OVER (PARTITION BY AT.Phase, AT.DailyUID) AS PhaseDuration,
            SUM(AT.Duration) OVER (PARTITION BY AT.DailyUID) AS TotalDuration
        FROM ActivityTable AT
    )
    SELECT S.Phase, S.DailyUID, S.PhaseDuration, S.TotalDuration, S.PhaseDuration * 1.0 / S.TotalDuration * 1000
    FROM Sums S

So, grouping is done in `Sums` CTE by phase and day first and then by day only.

[编辑 - 使其适用于 MySql]

SELECT S.Phase, S.DailyUID, S.PhaseDuration, S.TotalDuration, S.PhaseDuration * 1.0 / S.TotalDuration * 1000
FROM (
    SELECT DISTINCT ATOut.DailyUID, ATOut.Phase, PD.PhaseDuration, TD.TotalDuration
    FROM ActivityTable ATOut
        JOIN (SELECT AT.Phase, AT.DailyUID, SUM(AT.Duration) PhaseDuration FROM ActivityTable AT GROUP BY Phase, DailyUID) PD 
            ON PD.Phase = ATOut.Phase AND PD.DailyUID = ATOut.DailyUID
        JOIN (SELECT AT.DailyUID, SUM(AT.Duration) TotalDuration FROM ActivityTable AT GROUP BY DailyUID) TD 
            ON TD.DailyUID = ATOut.DailyUID
) S

如何创建 table + 计算成本涉及 2 列

How to create table + calculate cost involve 2 columns

mysql

calculated-columns