SQL Query to only pull one record for first day of month, for every month

Question

I have a table that has one record per day. Eg (this is just the date col of the table)

2018-07-08 03:00:00
2018-07-07 03:00:00
2018-07-06 03:00:00
2018-07-05 03:00:00
2018-07-04 03:00:00
2018-07-03 03:00:00
2018-07-02 03:00:00
2018-07-01 03:00:00
2018-06-30 03:00:00
2018-06-29 03:00:00

This data goes back a few years

I want to pull just the first day of month record, for all months in the table.

What is the SQL to do that?

(On SQL Server 2014)

Answer 1

You can use row_number() function :

select *
from (select *, row_number() over (partition by datepart(year, date), datepart(month, date) order by datepart(day, date)) seq
      from table
     ) t
where seq = 1;

Perhaps you also need year in partition clause.

Answer 2

If all your time are zeroed all you do need is to get everything where DATEPART is first day.

select * from dbo.MyTable mt where DATEPART(day, mt.MyDate) = 1

It will work if you got one row per day. Off course you will need to use DISTINCT or an aggregation if you got more than one row per day.

Answer 3

I would use the day() function:

select t.*
from t
where day(t.MyDate) = 1;

Neither this nor datepart() are ANSI/ISO-standard, but there are other databases that support day() . The standard function is extract(day from t.MyDate) .

If you want the first record in the table for each month -- but for some months, that might not be day 1 -- then you can use row_number() . One method is:

select top (1) with ties t.*
from t
order by row_number() over (partition by year(mydate), month(mydate) order by day(mydate) asc);

Answer 4

Though this has been answered, you can use date from parts in MS SQL as well.

  create table  #temp (dates date) 

 insert into #temp values  ('2018-01-02'),('2018-01-05'), ('2018-01-09'), ('2018-01-10')

 select * from #temp 

 dates
 2018-01-02
 2018-01-05
 2018-01-09
 2018-01-10

 You can use this to get beginning of the month 

     select DATEFROMPARTS(year(dates), month(dates), 01) Beginningofmonth  from #temp 
     group by DATEFROMPARTS(year(dates), month(dates), 01)

 Output: 
 Beginningofmonth
 2018-01-01