Question

查询：

SELECT   
SUM(TotalDue), DATENAME(dw, OrderDate), COUNT(DISTINCT ProductID)  
FROM Sales.SalesOrderHeader headers  
INNER JOIN Sales.SalesOrderDetail details  
ON  headers.SalesOrderID = details.SalesOrderID  
GROUP BY DATENAME(dw, OrderDate);

我目前的结果是：

560644420,8959  Wednesday   262
432484099,6362  Thursday    260
478917953,3263  Saturday    251
212585108,7496  Friday  238
457072471,4514  Monday  241
381773345,5659  Sunday  259
403492724,4161  Tuesday 235

现在我想调整这些数据，所以我有几天作为列。问题是我不知道如何实现这个结果 - 如何让第一行告诉它代表总和，其次是它计算不同的ID。

提前感谢任何建议和帮助

到目前为止我的尝试，也许这个查询会让你的回答更容易，因为你不必编写所有健壮的代码：

SELECT [Total], [Monday], [Tuesday]
FROM 
    (
        SELECT SUM(TotalDue)[Total], DATENAME(dw, OrderDate) [Day], COUNT(DISTINCT ProductID)[different products]
        FROM Sales.SalesOrderHeader headers
        INNER JOIN Sales.SalesOrderDetail details ON  headers.SalesOrderID = details.SalesOrderID
        GROUP BY DATENAME(dw, OrderDate)
) as source
PIVOT(
    COUNT([different products]) FOR [Day] IN ([Monday], [Tuesday])
) as pivoted;

我的预期结果（日期顺序并不重要）：

                        Monday            Tuesday         Wednesday ...
Sums of orders          457072471,4514    403492724,4161  560644420,8959
Different items count   241               235             262

Answer 1

你可以结合两个数据透视语句来做这件事。

Select  Totals,Monday  ,Tuesday from (
SELECT  'TotalSum' As Totals, [Monday] as Monday, [Tuesday]
FROM 
    (
        SELECT sum(Totaldue) As TotalDue, DATENAME(dw, OrderDate) [Day] 
        FROM Sales.SalesOrderHeader headers
        INNER JOIN Sales.SalesOrderDetail details ON  headers.SalesOrderID = details.SalesOrderID
        group by  DATENAME(dw, OrderDate) 


) as source
PIVOT(
    sum(TotalDue) FOR [Day] IN ([Monday], [Tuesday])
    ) as pivoted

    union all

SELECT  'TotalCount' As Totals, [Monday] as Monday, [Tuesday]
FROM 
    (
        SELECT  DATENAME(dw, OrderDate) [Day], count(distinct ProductID) as DistinctProduct
        FROM Sales.SalesOrderHeader headers
        INNER JOIN Sales.SalesOrderDetail details ON  headers.SalesOrderID = details.SalesOrderID
        group by  DATENAME(dw, OrderDate) 


) as source
PIVOT(
    sum(DistinctProduct) FOR [Day] IN ([Monday], [Tuesday]) 
    )as pivoted
    )  x

Answer 2

你不能一次做多个支点，所以你必须做两次：

;WITH Data AS
(
    SELECT SUM(TotalDue)[Total], DATENAME(dw, OrderDate) [Day], COUNT(DISTINCT ProductID)[different products]
    FROM 
        Sales.SalesOrderHeader headers
        INNER JOIN Sales.SalesOrderDetail details ON  headers.SalesOrderID = details.SalesOrderID
    GROUP BY 
        DATENAME(dw, OrderDate)
),
PrePivot1 AS
(
    SELECT
        D.Day,
        D.[different products]
    FROM
        Data AS D
),
PrePivot2 AS
(
    SELECT
        D.Day,
        D.[Total]
    FROM
        Data AS D
)
SELECT
    Concept = 'Different items count',
    pivoted.Monday,
    pivoted.Tuesday,
    pivoted.Wednesday,
    pivoted.Thursday,
    pivoted.Friday,
    pivoted.Saturday,
    pivoted.Sunday
FROM
    PrePivot1 AS D
    PIVOT (
        SUM([different products]) FOR [Day] IN ([Monday], [Tuesday], [Wednesday], [Thursday], [Friday], [Saturday], [Sunday])
    ) as pivoted

UNION ALL

SELECT
    Concept = 'Sums of orders',
    pivoted.Monday,
    pivoted.Tuesday,
    pivoted.Wednesday,
    pivoted.Thursday,
    pivoted.Friday,
    pivoted.Saturday,
    pivoted.Sunday
FROM
    PrePivot2 AS D
    PIVOT (
        SUM([Total]) FOR [Day] IN ([Monday], [Tuesday], [Wednesday], [Thursday], [Friday], [Saturday], [Sunday])
    ) as pivoted;

请注意，我将PIVOT汇总功能从COUNT()更改为SUM()，因为您似乎正在添加产品ID。

我正在做2个“PrePivots”的原因是PIVOT操作由GROUP BY中未引用的所有列执行隐式PIVOT，因此其他列（如Total）对于第一个数据透视表）将被分组，并且对于每个不同的总值，您将有一行。

转发麻烦，SQL Server

2 个答案: