Group By day of month across calendar days - SQL Server 2012 [duplicate] - sql

Question

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Efficiently convert rows to columns in sql server
(5 answers)
Closed 8 years ago.
I have a query that produces results like the following:
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
Here is the query:
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
I'm puzzling over how to get this a little differently. Notice how Amanda checked in 3 times in July, on the 3rd, the 9th, and the 23rd. I need the results to show the check-in count across every day of the month. So in place of the "Date" column I need 31 columns (for each possible day of the longest possible month), like so:
1 2 3 4 5 6 7 8 9 10 .....
==============================================
0 0 1 0 0 0 0 0 1 0 ......
Notice the "check-in" count under 3 and 9, for July 3rd and July 9th. How can I build the query to produce results like that?

Answer 1

Ugly answer...but you should be able to create 31 case statements to get this to work. In your select (at the end works) include:
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
Would be curious if there is a better answer, but it should function for you. They are sum columns, so the existing group bys should function fine.