选择 ID、计数 (ID) 和按日期分组

Question

我有一个包含 id 和日期（月/年）列的文章表，首先我想计算 id 并按日期对它们进行分组，然后我想查看哪个 id 属于单个日期组像这样查询：

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

我有 2 个查询

Select Count(id) 
from article 
group by date

和

Select id 
from article

给出结果；

count date            id  date
-------------         ----------
2     01/2015         1  01/2015
1     02/2015         2  01/2015 
4     03/2015         3  02/2015

我需要一个像

这样的查询

select count(id), id, date 
from....

它带来了在我的 C# 代码中使用的 id、count、date 列。

有人可以帮我解决这个问题吗？

Answer 1

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

Sql fiddle

Answer 2

不清楚您希望如何在结果中生成 id 字段。如果你 想要手动生成它然后使用RANK() 或者如果你想得到它 从表id 值中，您可以使用max() 或min()(取决于 根据您的预期结果）

使用RANK()Fiddle Demo Here

试试：

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)

查询：

 select count(id) as count,dt,RANK() 
    over(order by count(id)) as id from tt group by dt

EDIT2： 或者你可以使用MAX() 或MIN()

喜欢：

select count(id) as count,dt,Min(id) as id from tt group by dt

或

select count(id) as count,dt,MAX(id) as id from tt group by dt

Answer 3

不能在一个查询中完全做到这一点，但您可以使用 CTE 生成单个结果集：

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

结果：

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 4

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

有了这些数据：

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