使用 OVER 显示百分比

Question

有人可以告诉我一种使用 OVER 完成以下任务的方法吗？

这就是我所拥有的

JobId     AgentID
--------------------------------------------
Job1      1
Job1      8
Job1      8
Job2      6
Job2      6
Job3      5

我需要列出每个作业的顶级代理并显示该作业在该代理上运行的百分比：

JobId     AgentID     TopPercent
--------------------------------------------
Job1      8           66
Job2      6           100
Job3      5           100

这需要通过计算行数来完成，因为实际上有数百个工作和代理。

顺便说一句，我正在使用 SQL 2008。

Answer 1

你可以这样写……这里有一个例子：

create table test (jobid varchar(10), agentid int);

insert into test values
('job1',      1)
,('job1',      8)
,('job1',      8)
,('job2',      6)
,('job2',      6)
,('job3',      5);

-- using OVER
select distinct jobid, agentid, 
(count(*) over (partition by jobid, agentid))*100 /
(count(*) over (partition by jobid)) as TopPercent
from test

如果您不想使用 OVER，请尝试以下操作：

-- not using OVER
with 

jobcount as (
  select jobid, count(*) as totbyjob from test group by jobid
),

jobagentcount as (
  select jobid, agentid, count(*) as totbyjobagent
  from test
  group by jobid, agentid
)

select 
  jac.jobid, 
  jac.agentid, 
  totbyjobagent*100/totbyjob as TopPercent
from jobagentcount jac
inner join jobcount jc
  on jac.jobid = jc.jobid

示例 SQLFiddle：http://sqlfiddle.com/#!3/f867c/18

要按百分比只获得排名靠前的职位，这样的查询就足够了：

with percents as (
  select distinct jobid, agentid, 
  (count(*) over (partition by jobid, agentid))*100 /
  (count(*) over (partition by jobid)) as toppercent
  from test
)

select jobid, agentid, toppercent from (
  select 
    percents.*,
    row_number() over(partition by jobid order by toppercent desc) as rank
  from percents
) main
where rank = 1

示例如下：http://sqlfiddle.com/#!3/f867c/27