需要查询以按照以下方式获取数据。

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英文:

Need query to get data as per below

问题

Input Table是

Target| Agent| RankA
-|-|-
15000| A| null  
15000| A| null  
14500| A| null

所需输出是  

Target| Agent| RankA
-|-|-
15000| A| 1| 33  
15000| A| 1| 33  
14500| A| 3| 100  

因为第一条记录是15000,所以它的排名是1
因为第二条记录也是15000,所以它的排名也是1
因为第三条记录是14500,所以它的排名是3,而不是2,因为即使前两条记录相同,现在的计数是2,所以第三条记录的新排名将是3

相应的查询是

```sql
here is script for data preparation  
```sql
SELECT 15000 TARGET,
       '' AS AGENT,
       NULL RANKA
INTO #TABLEA;  

INSERT INTO #TABLEA
VALUES (1400, 'B', NULL);
英文:

Input Table is

Target Agent RankA
15000 A null
15000 A null
14500 A null

required output is

Target Agent RankA
15000 A 1
15000 A 1
14500 A 3

as first record is 15000, so its rank is 1
as second record is 15000, so its rank is also 1
as third record is 14500, so its rank is also 3 and not 2 because even though first two records have same record, the count is now 2 so new rank will be 3 for third record

need query for the same..

here is script for data preparation

SELECT 15000 TARGET,
       '' AS AGENT,
       NULL RANKA
INTO #TABLEA;  

INSERT INTO #TABLEA
VALUES (1400, 'B', NULL);

答案1

得分: 1

使用 rank() 来获取排名,以及 rank/count 来获取百分位:

with cte as (
  select Target, Agent, rank() over (partition by Agent order by Target desc) as Ranka
  from TableA
)
select Target, Agent, cast(Ranka as float)/CAST(count(*) over (partition by Agent) AS FLOAT)*100 as Percentile
from cte

或者,如果你不需要百分位:

select Target, Agent, rank() over (partition by Agent order by Target desc) as Ranka
from TableA

演示在此

英文:

You can do it using rank() to get the rank, and rank/count to get Percentile

with cte as (
  select Target, Agent, rank() over (partition by Agent order by Target desc) as Ranka
  from TableA
)
select Target, Agent, cast(Ranka as float)/CAST(count(*) over (partition by Agent) AS FLOAT)*100 as Percentile
from cte

Or this if you dont need Percentile :

select Target, Agent, rank() over (partition by Agent order by Target desc) as Ranka
from TableA

Demo here

huangapple
  • 本文由 发表于 2023年3月31日 18:11:01
  • 转载请务必保留本文链接:https://go.coder-hub.com/75897318.html
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