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我有一个程序，它为具有“工人”头衔的评估人员带来技能评级，他们必须携带分配给他们的文件编号。该程序还引入了每个工人所属的报告线。

SELECT distinct 
o.VP,
o.AVP,
o.Director,
o.Supervisor,
o.Worker,
bs.File_NBR,
s.Skill
bs.score
FROM    [New_EEs].[dbo].[SBC_Best_Scores] bs
inner join new_ees.dbo.SBC_Skills s
on   bs.Skill_NBR=s.SKILL_NBR
inner join gw_PPP.dbo.Org_Hierarchy oon 
bs.File_NBR=o.File_NBR;

我得到一个看起来像这样的数据集：

VP  AVP Director    Supervisor  Worker  File_NBR    Skill   Rating
Gerald  Kris    Doris   NULL    Mack    107812  B2  4
Gerald  Kris    Doris   NULL    Mack    107812  D1  3
Gerald  Kris    Doris   NULL    Mack    107812  D2  3
Gerald  Kris    Doris   NULL    Mack    107812  D3  3
Gerald  Kris    Doris   NULL    Mack    107812  E1  4
Gerald  Kris    Mike    NULL    Brady   109080  A1  5
Gerald  Kris    Mike    NULL    Brady   109080  B1  4
Gerald  Kris    Mike    NULL    Brady   109080  B2  3
Gerald  Kris    Mike    NULL    Brady   109080  B3  4
Gerald  Kris    Mike    NULL    Brady   109080  C1  4
Gerald  Kris    Mike    NULL    Brady   109080  C2  4
Gerald  Kris    Mike    NULL    Brady   109080  C3  0
Kim Harry   NULL    Grant   Tom 108457  B1  4
Kim Harry   NULL    Grant   Tom 108457  B2  4
Kim Harry   NULL    Grant   Tom 108457  C1  4
Kim Harry   NULL    Grant   Tom 108457  C2: 5
Kim Harry   NULL    Grant   Tom 108457  C5  5
Kim Harry   NULL    Grant   Tom 108457  D1  4
Kim Harry   NULL    Grant   Tom 108457  D2  5
Kim Harry   NULL    Grant   Tom 108457  D3  4
Kim Harry   NULL    Grant   Jean    106934  C5  4
Kim Harry   NULL    Grant   Jean    106934  D1  5
Kim Harry   NULL    Grant   Jean    106934  D3  5
Kim Harry   NULL    Grant   Raphe   108901  B2  5
Kim Harry   NULL    Grant   Raphe   108901  C2  5
Kim Harry   NULL    Grant   Raphe   108901  C3  4
Kim Harry   NULL    Grant   Raphe   108901  C5  5
Kim Harry   NULL    Grant   Raphe   108901  D2  5
Kim Harry   NULL    Grant   Raphe   108901  E1  5
Kim Harry   NULL    Grant   Tyika   107923  B1  5
Kim Harry   NULL    Grant   Tyika   107923  B2  5
Kim Harry   NULL    Grant   Tyika   107923  D2  4
Kim Harry   NULL    Grant   Tyika   107923  D3  4

评级级别为 1 到 5。我需要做的是创建一个表格，显示按 Vp、AVP、主管和主管分组的每种技能对工人的每个评级的计数和百分比。所以所有最终都在 AVP 之下的作品和所有最终在导演之下的作品等等。

Name    Role    Skill   Count of    % of    Count of      % of  
                              Rating 1   Rating 1  Rating 2   Rating 2
Gerald  VP  A1  100 29% 130 33%
Gerald  VP  B1  95  28% 95  24%
Gerald  VP  B2  120 35% 70  18%
Gerald  VP  B3  30  9%  100 25%
Kim VP  A1              
Kim VP  B1              
Kim VP  B2      and so on       
Kim VP  B3              
Kris    AVP A1              
Kris    AVP B1              
Kris    AVP B2              
Kris    AVP B3              
Harry   AVP A1              
Harry   AVP B1              
Harry   AVP B2              
Harry   AVP B3              
Doris   Director    A1              
Doris   Director    B1              
Doris   Director    B2              
Doris   Director    B3              
Mike    Director    A1              
Mike    Director    B1              
Mike    Director    B2              
Mike    Director    B3              
Grant   Supervisor  A1              
Grant   Supervisor  B1              
Grant   Supervisor  B2              
Grant   Supervisor  B3

任何帮助都会很棒！谢谢！

4

1 回答 1

1

由于您在不同的列中具有不同的角色，因此要获得紧凑的查询，您需要动态 sql 或复杂的数据透视。因此，我选择仅复制和粘贴，因为我认为复杂性不值得您拥有 4 个角色。

我已将您的查询命名为 T 作为示例。

with roles as (
    select VP as Name, 'VP' as Role, Skill, Rating from t where VP is not null
  union all 
    select AVP as Name, 'AVP' as Role, Skill, Rating from t where AVP is not null
  union all 
    select Director as Name, 'Director' as Role, Skill, Rating from t where Director is not null
  union all 
    select Supervisor as Name, 'Supervisor' as Role, Skill, Rating from t where Supervisor is not null
), counts as (
  select Name, Role, Skill
      ,count(case when rating = 1 then 1 else NULL end) as [Count of Rating 1]
      ,count(case when rating = 2 then 1 else NULL end) as [Count of Rating 2]
      ,count(case when rating = 3 then 1 else NULL end) as [Count of Rating 3]
      ,count(case when rating = 4 then 1 else NULL end) as [Count of Rating 4]
      ,count(case when rating = 5 then 1 else NULL end) as [Count of Rating 5]
      ,count(*) as TotalCount
    from roles
    group by Name, Role, skill
)
select Name, Role, Skill
,[Count of Rating 1]
,CONVERT(varchar(10), convert(int,100.0 * [Count of Rating 1]/NULLIF(TotalCount, 0))) + '%' as [% of Rating 1]
,[Count of Rating 2]
,CONVERT(varchar(10), convert(int,100.0 * [Count of Rating 2]/NULLIF(TotalCount, 0))) + '%' as [% of Rating 2]
,[Count of Rating 3]
,CONVERT(varchar(10), convert(int,100.0 * [Count of Rating 3]/NULLIF(TotalCount, 0))) + '%' as [% of Rating 3]
,[Count of Rating 4]
,CONVERT(varchar(10), convert(int,100.0 * [Count of Rating 4]/NULLIF(TotalCount, 0))) + '%' as [% of Rating 4]
,[Count of Rating 5]
,CONVERT(varchar(10), convert(int,100.0 * [Count of Rating 5]/NULLIF(TotalCount, 0))) + '%' as [% of Rating 5]
from counts
order by Name, skill

我在这里所做的是将所有角色联合在一起，对角色名称进行硬编码。roles重新组织表格，以便拥有 VP 的每个人都获得该 VP 的一行，拥有 AVP 的每个人都获得该 AVP 的一行，...counts然后计算每个名称、角色和技能的所有工人。最后的选择计算百分比。

这是一个展示它的小提琴：http ://sqlfiddle.com/#!3/fe09d/15

于 2013-07-23T23:02:43.777 回答