Question

我有一个查询，它返回所有发货部件，工厂部件，每个部件的输入日期以及各自的电子邮件。但是我需要每个人输入的所有船舶部件和工厂部件的总计数，所以我添加了你在下面看到的COUNT函数，但我不断得到这个错误＆＃34;每个GROUP BY表达式必须包含至少一个不是外部参考。＆＃34;有什么想法吗？

select Distinct cc.FirstName+' '+cc.LastName AS 'Name', UEmail AS 'Email', sp.SEntered AS 'Date Entered', COUNT(sp.SHIPID) AS 'Shipments',   COUNT(pn.PNumber) AS 'Factory Parts'
from R_USERS_Detail U
join R_SHIPMENTS sp on sp.UID=U.UID
join R_SHIPMENTOrders sd on sd.SHIPID=sp.SHIPID
join R_PARTOrders pn on pn.PNumberID=sd.PNumberID
join R_Clients c on c.COClientID=sp.COClientID AND c.COClientAcronym LIKE '%AME%'
join R_Contacts cc on cc.COContactID=u.COContactID
group by u.UEmail, sp.SEntered, 'Name'

Answer 1

您必须使用group by claus中的确切表达式，而不是别名。

select cc.FirstName+' '+cc.LastName AS 'Name'
     , UEmail AS 'Email'
     , sp.SEntered AS 'Date Entered'
     , COUNT(sp.SHIPID) AS 'Shipments'
     , COUNT(pn.PNumber) AS 'Factory Parts'
from R_USERS_Detail U
join R_SHIPMENTS sp on sp.UID=U.UID
join R_SHIPMENTOrders sd on sd.SHIPID=sp.SHIPID
join R_PARTOrders pn on pn.PNumberID=sd.PNumberID
join R_Clients c on c.COClientID=sp.COClientID AND c.COClientAcronym LIKE '%AME%'
join R_Contacts cc on cc.COContactID=u.COContactID
group by u.UEmail, sp.SEntered, cc.FirstName+' '+cc.LastName

使用多个COUNTS（）和GROUP BY进行查询

1 个答案: