Question

我有一个查询产生如下结果：

Company     Member          Account     Date        Check-ins
================================================================
Acme, Inc.  Amanda Smith    4145886     7/3/2014    1
Acme, Inc.  Amanda Smith    4145886     7/9/2014    1
Acme, Inc.  Amanda Smith    4145886     7/23/2014   1
Acme, Inc.  Gladys Jones    800138618   7/5/2014    1
Acme, Inc.  Joe Ortega      800123972   7/15/2014   1
Acme, Inc.  Joe Ortega      800123972   7/29/2014   1

以下是查询：

Select 
    com.CompanyName as [Company], 
    p.FirstName + ' ' + p.LastName as [Member],
    a.AccountID as [Account], 
    CAST(mc.CheckInDate AS Date) [Date], 
    count(*) AS [Check-ins]
from 
    gym.Person p 
join 
    gym.AccountPeople ap on p.PersonID = ap.PersonID
join 
    gym.Account a on a.AccountID = ap.AccountID
join 
    gym.MembershipStatus ms on a.MembershipStatusID = ms.MembershipStatusID 
join 
    gym.Company com on a.CompanyID = com.CompanyID
join 
    gym.MemberCheckin mc on mc.PersonID = p.PersonID
where 
    mc.CheckInDate > '2014-7-1'
    and mc.CheckInDate < DATEADD(dd, 1, CAST('2014-7-31' AS Date))
GROUP BY 
    com.CompanyName, a.AccountID, 
    p.FirstName + ' ' + p.LastName, CAST(mc.CheckInDate AS Date)
order by 
    com.CompanyName, p.FirstName + ' ' + p.LastName

我对如何以不同的方式解决这个问题感到困惑。请注意Amanda在7月，3日，9日和23日进行了3次检查。我需要结果显示每月每天的登记入住时间。因此，取代“日期”列，我需要31列（对于最长可能月份的每一天），如下所示：

1   2   3   4   5   6   7   8   9   10 .....
==============================================
0   0   1   0   0   0   0   0   1   0 ......

注意7月3日和7月9日3和9下的“办理登机手续”计数。如何构建查询以生成类似的结果？

Answer 1

丑陋的答案......但你应该能够创建31个案例陈述来使其发挥作用。在你的选择（最后的作品）包括：

sum(case when day([date]) = 1 then 1 else 0 end) as day1,
sum(case when day([date]) = 2 then 1 else 0 end) as day2,
sum(case when day([date]) = 3 then 1 else 0 end) as day3,
sum(case when day([date]) = 4 then 1 else 0 end) as day4,
etc
sum(case when day([date]) = 31 then 1 else 0 end) as day31

如果有更好的答案会很好奇，但它应该适合你。它们是sum列，因此现有的group by应该可以正常运行。

分组跨日历日的每月日期 - SQL Server 2012

1 个答案: