Question

与我上一篇文章类似，但深入挖掘一下。本报告的下一个任务是列出每个位置的前10位客户 - 例如：00468,00469,00471等的前10位客户

select top 10 T1.LocationID, T1.CustName, T1.Quantity--, t1.Salesperson
from
(
select 
     SUM(Tkscale.Qty)Quantity,
        Slcust.Name CustName,
        Slperson.Name Salesperson,
        Inlocat.LocationID LocationID,
        Inlocat.Description Location
    from 
Tkscale
      left outer join Slcust on Tkscale.CustomerID = Slcust.CustomerID 
      left outer join Slperson on slcust.SalespersonID = Slperson.SalespersonID
      left outer join Inlocat on Inlocat.LocationID = Tkscale.LocationID
     where 
slperson.Name like 'Tammy%'
   group by 
Inlocat.LocationID, inlocat.description, Tkscale.CustomerID, Slcust.Name, Slperson.Name
) T1
order by  T1.Quantity desc, T1.LocationID, T1.CustName

输出：

Answer 1

使用row_number

SELECT  T1.LocationID
    ,T1.CustName
    ,T1.Quantity --, t1.Salesperson
FROM (
    SELECT SUM(Tkscale.Qty) Quantity
        ,Slcust.NAME CustName
        ,Slperson.NAME Salesperson
        ,Inlocat.LocationID LocationID
        ,Inlocat.Description Location
        ,row_number() over (partition by Inlocat.LocationID order by SUM(Tkscale.Qty) desc) rowid 
    FROM Tkscale
    LEFT JOIN Slcust
        ON Tkscale.CustomerID = Slcust.CustomerID
    LEFT JOIN Slperson
        ON slcust.SalespersonID = Slperson.SalespersonID
    LEFT JOIN Inlocat
        ON Inlocat.LocationID = Tkscale.LocationID
    WHERE slperson.NAME LIKE 'Tammy%' 
    GROUP BY Inlocat.LocationID
        ,inlocat.description
        ,Tkscale.CustomerID
        ,Slcust.NAME
        ,Slperson.NAME
    ) T1
where t1.rowid < 11
ORDER BY T1.LocationID
    ,T1.rowid desc

Answer 2

select T1.LocationID, T1.CustName, T1.Quantity--, t1.Salesperson
from
(
select 
     SUM(Tkscale.Qty)Quantity,
        Slcust.Name CustName,
        Slperson.Name Salesperson,
        Inlocat.LocationID LocationID,
        Inlocat.Description Location
       ,NoGapsWithTies = DENSE_RANK() OVER (PARTITION BY Inlocat.LocationID ORDER BY SUM(Tkscale.Qty) DESC)
       ,WithGapsAndTies = RANK() OVER (PARTITION BY Inlocat.LocationID ORDER BY SUM(Tkscale.Qty) DESC)
       ,NoGapsNoTiesAbitraryChoiceIfTie = ROW_NUMBER() OVER (PARTITION BY Inlocat.LocationID ORDER BY SUM(Tkscale.Qty) DESC)
    from 
Tkscale
      left outer join Slcust on Tkscale.CustomerID = Slcust.CustomerID 
      left outer join Slperson on slcust.SalespersonID = Slperson.SalespersonID
      left outer join Inlocat on Inlocat.LocationID = Tkscale.LocationID
     where 
slperson.Name like 'Tammy%'
   group by 
Inlocat.LocationID, inlocat.description, Tkscale.CustomerID, Slcust.Name, Slperson.Name
) T1
WHERE
    --NoGapsWithTies <=10
    WithGapsAndTies <= 10
    --NoGapsNoTiesAbitraryChoiceIfTie <=10
order by  T1.Quantity desc, T1.LocationID, T1.CustName

这是有效的，因为在计算窗函数之前，结果集是由group by构建的。使用DENSE_RANK会给你带来无间隙的关系，RANK会给你带来联系和差距，ROW_NUMBER会给你没有联系，没有差距，但任意选择什么时候他们是客户之间的关系

客户和位置排名前10位

2 个答案: