Select ID, Count(ID) and Group by Date - sql

Question

I have an article table which has id and date (month/year) columns, first of all I would like to count ids and group them by date, then I would like to see which id belongs to which date group in single query like that:
id date count
-----------------
1 01/2015 2
2 01/2015 2
3 02/2015 1
4 03/2015 4
5 03/2015 4
6 03/2015 4
7 03/2015 4
I have 2 queries
Select Count(id)
from article
group by date
and
Select id
from article
gives results;
count date id date
------------- ----------
2 01/2015 1 01/2015
1 02/2015 2 01/2015
4 03/2015 3 02/2015
I need a single query like
select count(id), id, date
from....
which brings id, count, date columns to use in my C# code.
Can someone help me with this?

Answer 1

SELECT id,
date,
COUNT(*) OVER (PARTITION BY date) AS Count
FROM article
Sql fiddle

Answer 2

Can't quite do that in one query, but you could use a CTE to produce a single result set:
create table #tt (id int null, dt varchar(8))
insert #tt values
(1,'01/2015'),
(2,'01/2015'),
(3,'02/2015'),
(4,'03/2015'),
(5,'03/2015'),
(6,'03/2015'),
(7,'03/2015')
;with cteCount(d, c) AS
(
select dt, count(id) from #tt group by dt
)
select id, dt, c
from #tt a
inner join cteCount cc
on a.dt = cc.d
drop table #tt
results:
id dt c
1 01/2015 2
2 01/2015 2
3 02/2015 1
4 03/2015 4
5 03/2015 4
6 03/2015 4
7 03/2015 4

Answer 3

if not exists(select * from TEST.sys.objects where type=N'U' and name=N'article')
begin
create table article(
[id] int,
[date] date)
end
with this data:
insert into article(id,date) values(1,convert(date,'15/01/2015',103));
insert into article(id,date) values(1,convert(date,'15/02/2015',103));
insert into article(id,date) values(2,convert(date,'15/03/2015',103));
insert into article(id,date) values(2,convert(date,'15/01/2015',103));
insert into article(id,date) values(3,convert(date,'15/02/2015',103));
insert into article(id,date) values(4,convert(date,'15/03/2015',103));
insert into article(id,date) values(5,convert(date,'15/01/2015',103));
insert into article(id,date) values(5,convert(date,'15/02/2015',103));
insert into article(id,date) values(1,convert(date,'15/03/2015',103));
insert into article(id,date) values(2,convert(date,'15/01/2015',103));
insert into article(id,date) values(3,convert(date,'15/02/2015',103));
insert into article(id,date) values(4,convert(date,'15/03/2015',103));
insert into article(id,date) values(5,convert(date,'15/01/2015',103));
insert into article(id,date) values(1,convert(date,'15/02/2015',103));
insert into article(id,date) values(2,convert(date,'15/03/2015',103));
insert into article(id,date) values(3,convert(date,'15/01/2015',103));
insert into article(id,date) values(4,convert(date,'15/03/2015',103));
select id,[date], count(id) [count] from article
group by [date],[id]
the result:
id date count
1 2015-01-15 1
1 2015-02-15 2
1 2015-03-15 1
2 2015-01-15 2
2 2015-03-15 2
3 2015-01-15 1
3 2015-02-15 2
4 2015-03-15 3
5 2015-01-15 2
5 2015-02-15 1

Answer 4

It's not clear how you want to generate the id field in result. if you
want to generate it manually then use RANK() or if you want to get it
from the table id value then you can use max() or min()(depends
upon on your expected result)
Use RANK() Fiddle Demo Here
Try:
create table tt (id int null, dt varchar(8),count int)
insert tt values
(1,'01/2015',2),
(2,'01/2015',2),
(3,'02/2015',1),
(4,'03/2015',4),
(5,'03/2015',4),
(6,'03/2015',4),
(7,'03/2015',4)
Query:
select count(id) as count,dt,RANK()
over(order by count(id)) as id from tt group by dt
EDIT2:
or you just can use MAX() or MIN()
like:
select count(id) as count,dt,Min(id) as id from tt group by dt
or
select count(id) as count,dt,MAX(id) as id from tt group by dt