2

我有四列要排名。它们需要按员工 ID 分组，然后按订单号从低到高列出。然后当一切都井井有条时，我真的想得到城市在该顺序中的排名。如果同一名员工的同一个城市一个接一个地列出，那么我希望那些排名相同。

下表的示例如下。顺序是正确的，但排名不是我想要做的。

Name            Employee_ID     Order_Number     City       Rank   
John               1                1            Boston       1  
John               1                2            Boston       2  
Will               2                1            Peabody      1  
Will               2                2            Weston       2   
Will               2                3            Newton       3



select Name, Employee_ID, Order_Number, City,
dense_rank() over(partition by Employee_ID order by Order_Number) as rank
from #Employee

我实际上想要的结果是：

Name            Employee_ID     Order_Number     City       Rank   
John               1                1            Boston       1  
John               1                2            Boston       1  
Will               2                1            Boston       1  
Will               2                2            Weston       2   
Will               2                3            Newton       3

然后我最终会删除重复的城市，最终得到： 

Name            Employee_ID     Order_Number     City       Rank   
John               1                1            Boston       1  
Will               2                1            Boston       1  
Will               2                2            Weston       2   
Will               2                3            Newton       3
4

2 回答 2

0

您可以尝试以下脚本来获得所需的输出。

SELECT Name, Employee_ID, Order_Number, City ,
ROW_NUMBER() OVER (PARTITION BY Employee_ID ORDER BY Order_Number) rank
(
    select Name, Employee_ID, Order_Number, City,
    dense_rank() over(partition by Employee_ID,city order by Order_Number) as rank
    from #Employee
)A
WHERE rank = 1

结果集的输出是-

Name    Employee_ID Order_Number    City    rank
John    1           1               Boston  1
Will    2           1               Peabody 1
Will    2           2               Weston  2
Will    2           3               Newton  3

检查Fiddle上的脚本输出。

于 2019-06-13T17:18:56.703 回答
0

您可以使用LAG()检查之前的城市是否相同。如果前一个城市不同或为空，那么我们就按原样进行排名，如果城市相同，那么排名 - 1 给我们与上面行相同的数字。演示

with cte as (select Name, Employee_ID, Order_Number, City,
dense_rank() over (partition by Employee_ID order by Order_Number) as rank,
lag(City) over (partition by Employee_ID order by Order_Number) as previousCity
from #Employee)
select 
    Name, Employee_ID, Order_Number, City,
    case when previousCity = city then rank - 1
         else rank end as rank
from cte
于 2019-06-13T17:22:43.720 回答