Recursive first day of each month for current getdate - sql-server

Question

Using T-SQL, I want a new column that will show me the first day of each month, for the current year of getdate().
After that I need to count the rows on this specific date. Should I do it with CTE or a temp table?

Answer 1

If 2012+, you can use DateFromParts()
To Get a List of Dates
Select D = DateFromParts(Year(GetDate()),N,1)
From (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12)) N(N)
Returns
D
2017-01-01
2017-02-01
2017-03-01
2017-04-01
2017-05-01
2017-06-01
2017-07-01
2017-08-01
2017-09-01
2017-10-01
2017-11-01
2017-12-01
Edit For Trans Count
To get Transactions (assuming by month). It becomes a small matter of a left join to created Dates
-- This is Just a Sample Table Variable for Demonstration.
-- Remove this and Use your actual Transaction Table
--------------------------------------------------------------
Declare #Transactions table (TransDate date,MoreFields int)
Insert Into #Transactions values
('2017-02-18',6)
,('2017-02-19',9)
,('2017-03-05',5)
Select TransMonth = A.MthBeg
,TransCount = count(B.TransDate)
From (
Select MthBeg = DateFromParts(Year(GetDate()),N,1)
,MthEnd = EOMonth(DateFromParts(Year(GetDate()),N,1))
From (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12)) N(N)
) A
Left Join #Transactions B on TransDate between MthBeg and MthEnd
Group By A.MthBeg
Returns
TransMonth TransCount
2017-01-01 0
2017-02-01 2
2017-03-01 1
2017-04-01 0
2017-05-01 0
2017-06-01 0
2017-07-01 0
2017-08-01 0
2017-09-01 0
2017-10-01 0
2017-11-01 0
2017-12-01 0

Answer 2

For an adhoc table of months for a given year:
declare #year date = dateadd(year,datediff(year,0,getdate() ),0)
;with Months as (
select
MonthStart=dateadd(month,n,#year)
from (values(0),(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11)) t(n)
)
select MonthStart
from Months
rextester demo: http://rextester.com/POKPM51023
returns:
+------------+
| MonthStart |
+------------+
| 2017-01-01 |
| 2017-02-01 |
| 2017-03-01 |
| 2017-04-01 |
| 2017-05-01 |
| 2017-06-01 |
| 2017-07-01 |
| 2017-08-01 |
| 2017-09-01 |
| 2017-10-01 |
| 2017-11-01 |
| 2017-12-01 |
+------------+
The first part: dateadd(year,datediff(year,0,getdate() ),0) adds the number of years since 1900-01-01 to the date 1900-01-01. So it will return the first date of the year. You can also swap year for other levels of truncation: year, quarter, month, day, hour, minute, second, et cetera.
The second part uses a common table expression and the table value constructor (values (...),(...)) to source numbers 0-11, which are added as months to the start of the year.

Answer 3

Not sure why you require recursive... But for first day of month you can try query like below:
Select Dateadd(day,1,eomonth(Dateadd(month, -1,getdate())))

Answer 4

declare #year date = dateadd(year,datediff(year,0,getdate() ),0)
;WITH months(MonthNumber) AS
(
SELECT 0
UNION ALL
SELECT MonthNumber+1
FROM months
WHERE MonthNumber < 11
)
select dateadd(month,MonthNumber,#year)
from months