Question

我使用Microsoft SQL Server中的AventureWorks2014数据库。我需要创建一个视图，显示CustomerID，全名和通过网络销售给客户的总金额。

我的问题是，我似乎无法获得与单个客户相对应的值，以便单个客户在我的结果中回答单行。这是我的代码，任何帮助将不胜感激。我基本上需要在网上显示销售给客户的总金额。

if object_id('vTotalWebSalesPerCustomer', 'v') is not null
    drop view vTotalWebSalesPerCustomer;
go

create view vTotalWebSalesPerCustomer
as
    select distinct 
        c.CustomerID,
        ltrim(rtrim(concat(concat(concat(concat(concat(concat(concat(p.Title, ' '), p.LastName), ', '), ' '), p.FirstName), ' '), p.Suffix))) as NomClient,
        soh.TotalDue
    from
        [Sales].[Customer] as c
    left join 
        [Person].[Person] as p on c.CustomerID = p.BusinessEntityID
    left join 
        [Sales].[SalesOrderHeader] as soh on soh.CustomerID = c.CustomerID
    where 
        year(soh.OrderDate) = 2014 
        and datepart(quarter, soh.OrderDate) = 1 
        and [OnlineOrderFlag] = 1
go

select * 
from vTotalWebSalesPerCustomer

由于

Answer 1

您可以使用deploymentResult上的VIEW使用以下GROUP BY：

SELECT

注意：您只需要一个CONCAT函数来连接所有字符串值。

Answer 2

DISTINCT过滤结果行以删除重复项。我认为你想要的是聚合行。也许这会做你想要的：

create view vTotalWebSalesPerCustomer as
    select      c.CustomerID,
                        ltrim(rtrim(concat(concat(concat(concat(concat(concat(concat(p.Title, ' '), p.LastName), ', '), ' '), p.FirstName), ' '), p.Suffix))) as NomClient,
                        sum(soh.TotalDue) as TotalDue
    from [Sales].[Customer] as c
    left join [Person].[Person] as p on c.CustomerID = p.BusinessEntityID
    left join [Sales].[SalesOrderHeader] as soh on soh.CustomerID = c.CustomerID
    where year(soh.OrderDate) = 2014 and datepart(quarter, soh.OrderDate)=1 and [OnlineOrderFlag] = 1
    group by c.CustomerID,NomClient

请注意，我删除了您想要合计金额的不同的sum运算符，并按客户ID和名称字段进行分组（SQL Server要求您在GROUP BY所有结果列中列出没有汇总的。）

Answer 3

听起来你需要使用GROUP BY和SUM（），例如：

SELECT column-names
 FROM table-name
WHERE condition
GROUP BY column-names
ORDER BY column-names

如：

SELECT SUM(O.TotalPrice), C.FirstName, C.LastName
  FROM [Order] O JOIN Customer C 
    ON O.CustomerId = C.Id
 GROUP BY C.FirstName, C.LastName
 ORDER BY SUM(O.TotalPrice) DESC

会回来：

Sum         FirstName   LastName
117483.39   Horst       Kloss
115673.39   Jose        Pavarotti
113236.68   Roland      Mendel
57317.39    Patricia    McKenna
52245.90    Paula       Wilson
34101.15    Mario       Pontes
32555.55    Maria       Larsson

如何消除答案中的重复值

3 个答案: