【Transact-SQL】统计某字段中的值第一次出现后的2小时内出现的次数

原文:【Transact-SQL】统计某字段中的值第一次出现后的2小时内出现的次数

table1
name               createdate
a              2011-03-01 10:00:00
a              2011-03-01 11:00:00
a              2011-03-01 14:00:00
b              2011-03-01 13:00:00
b              2011-03-01 13:20:00
b              2011-03-01 14:00:00

查询结果为
name             createdate                    count
a             2011-03-01 10:00:00               2
a             2011-03-01 14:00:00               1
b             2011-03-01 13:00:00               3

就相当于是统计name字段中的值在第一次出现后的2小时内，总共出现了几次？

这个是网上的解答：

declare @table1 table(name nvarchar,createdate smalldatetime)

insert into @table1
select ‘a‘,‘2011-03-01 10:00:00‘
union all select ‘a‘,‘2011-03-01 11:00:00‘
union all select ‘a‘,‘2011-03-01 14:00:00‘
union all select ‘b‘,‘2011-03-01 13:00:00‘
union all select ‘b‘,‘2011-03-01 13:20:00‘
union all select ‘b‘,‘2011-03-01 14:00:00‘

select name,
createdate,
(select count(createdate)
from @table1 b 
where a.name=b.name and 
a.createdate<=b.createdate and 
dateadd(hh,2,a.createdate) >= b.createdate
) as count

from @table1 a 
where not exists 
(select 1 from 
@table1 b 
where a.name=b.name and 
a.createdate>b.createdate and 
a.createdate<dateadd(hh,2,b.createdate)) 
group by name,createdate

但是这个解答其实是有问题的，当把临时表中的第3条数据的createdate改为‘2011-03-01 12:00:00‘，那么显示的结果是：

name        createdate                           count

a                2011-03-01 10:00:00        3

b                2011-03-01 13:00:00        3

在其中没有包括createdate为‘2011-03-01 12:00:00‘的记录，因为这个时间到为‘2011-03-01 10:00:00‘是超过2个小时了，也就是说为‘2011-03-01 10:00:00‘是第一个出现时间，到为‘2011-03-01 11:59:59‘为止，接下来应该是从‘2011-03-01 12:00:00‘开始的下个区间了，而这里显然是有问题的。

以下是我写的解法，虽然效率不是太高，但是能解决这个问题：

declare @table1 table(name nvarchar,createdate smalldatetime)

insert into @table1
select ‘a‘,‘2011-03-01 10:00:00‘
union all select ‘a‘,‘2011-03-01 11:00:00‘
union all select ‘a‘,‘2011-03-01 12:00:00‘
union all select ‘b‘,‘2011-03-01 13:10:00‘
union all select ‘b‘,‘2011-03-01 13:20:00‘
union all select ‘b‘,‘2011-03-01 14:30:00‘
union all select ‘b‘,‘2011-03-01 15:15:00‘
union all select ‘b‘,‘2011-03-01 16:00:00‘
union all select ‘b‘,‘2011-03-01 17:00:00‘



;with aa  --按照name分区，同时按照createdate排序编号
as
(
select name,
createdate,
ROW_NUMBER() over(partition by name
order by createdate) as k1
from @table1 
),

r
as
(
select v.name,
starts=v.k1,  --区间开始的编号

ends=isnull(
min(case when v.k1<a.k1 
and DATEADD(hour,2,v.createdate) <= a.createdate 
then a.k1 
else null 
end)-1,

max(case when v.k1<a.k1 
then a.k1 
else v.k1 
end)
),     --区间结尾的编号      

isnull(
min(case when v.k1<a.k1 
and DATEADD(hour,2,v.createdate) <= a.createdate 
then a.k1 
else null 
end)-1,

max(case when v.k1<a.k1 
then a.k1 
else v.k1 
end)
) - v.k1 as diff     --区间结尾编号与区间开始编号之间的差值

from aa v  
inner join aa a 
on v.name = a.name  --只关联name相等的
group by v.name,
v.k1
having isnull(
min(case when v.k1<a.k1 
and DATEADD(hour,2,v.createdate) <= a.createdate 
then a.k1 
else null 
end)-1,

max(case when v.k1<a.k1 
then a.k1 
else v.k1 
end)
)  >=v.k1   
and            
isnull(
max(case when v.k1>a.k1 and 
DATEADD(hour,-2,v.createdate) >= a.createdate
then v.k1 - 1 
else null 
end) + 1,

min(case when v.k1>a.k1 
then a.k1 
else v.k1 
end)
) = v.k1
)

--select * from r


select aa.name,
aa.createdate,
diff + 1
from r
inner join aa 
on aa.name = r.name
and aa.k1 =r.starts
where not exists
(select 1
from r rr
where rr.name = r.name and
rr.starts <> r.starts and
rr.starts < r.starts and
(rr.ends = r.ends or
rr.ends = r.starts)
)           

不过发现我的这个解法也是有问题的，最大的问题在与不能准确的确定上限在那里，还得继续考虑问题的解法。

下面这个也是有问题的：

declare @table1 table(name nvarchar,createdate smalldatetime)

insert into @table1
select ‘a‘,‘2011-03-01 10:00:00‘
union all select ‘a‘,‘2011-03-01 11:00:00‘
union all select ‘a‘,‘2011-03-01 12:00:00‘

--union all select ‘b‘,‘2011-03-01 13:10:00‘
--union all select ‘b‘,‘2011-03-01 13:20:00‘
--union all select ‘b‘,‘2011-03-01 14:30:00‘
--union all select ‘b‘,‘2011-03-01 15:15:00‘
--union all select ‘b‘,‘2011-03-01 16:00:00‘
--union all select ‘b‘,‘2011-03-01 17:15:00‘



;with a  --按照name分区，同时按照createdate排序编号
as
(
select name,
createdate,
ROW_NUMBER() over(partition by name
order by createdate) as k1
from @table1 
),

c
as
(
select a1.name,
a1.createdate,
a1.k1,
MIN(a2.createdate) as nextCreatedate,
MIN(a2.k1) as nextK1
from a a1
inner join a a2
on a1.name = a2.name
and a2.createdate < DATEADD(hour,2,a1.createdate)
and a2.createdate >= a1.createdate
and a1.k1 <= a2.k1
group by a1.name,
a1.createdate,
a1.k1
),

w
as
(
select name,
createdate,
k1

--null,
--null,
--null
from a
where k1 = 1

union all

select c.name,
c.nextCreatedate,
c.nextK1

from W
inner join a
on a.name = w.name
and dateadd(hour,2,w.createdate) <= a.createdate
and w.k1 <= a.k1
and w.createdate <> ‘2011-03-01 12:00:00‘ 
inner join c
on w.name = c.name 
and w.createdate = c.createdate
and w.k1 = c.k1
where w.k1 <=3
)

SELECT * 
FROM w

下面的解法是正确的，不过用的是T-SQL，不是纯sql了：

declare @table1 table(name nvarchar(100),createdate smalldatetime)

declare @table2 table(name nvarchar(100),createdate smalldatetime,rnum bigint)

declare @temp table(name nvarchar(100),createdate smalldatetime,rnum bigint)

declare @i int = 1;

insert into @table1
select ‘a‘,‘2011-03-01 10:00:00‘
union all select ‘a‘,‘2011-03-01 11:00:00‘
union all select ‘a‘,‘2011-03-01 12:00:00‘

union all select ‘b‘,‘2011-03-01 13:10:00‘
union all select ‘b‘,‘2011-03-01 13:20:00‘
union all select ‘b‘,‘2011-03-01 14:30:00‘
union all select ‘b‘,‘2011-03-01 15:15:00‘
union all select ‘b‘,‘2011-03-01 16:00:00‘
union all select ‘b‘,‘2011-03-01 17:16:00‘
union all select ‘b‘,‘2011-03-01 17:15:00‘


;with a  --按照name分区，同时按照createdate排序编号
as
(
select name,
createdate,
ROW_NUMBER() over(partition by name
order by createdate) as k1
from @table1 
)

insert into @table2
select * from a

insert into @temp
select name,
createdate,
rnum
from @table2
where rnum = 1

--select * from @temp


while @i <= (select MAX(rnum) from @table2)
begin
insert into @temp

select t2.name,
min(t2.createdate),
@i +1	
from @temp t1
inner join @table2 t2        
on t1.name = t2.name
and t2.createdate >= dateadd(hour,2,t1.createdate)	           
where t1.rnum = @i     
group by t2.name

set @i = @i + 1	
end

;with r
as
(
select name,
createdate
from @temp
group by name,
createdate
)

select r.name,
r.createdate,
COUNT(1)
from r
inner join @table1 t
on t.name = r.name
and t.createdate >= r.createdate
and t.createdate <DATEADD(HOUR,2,r.createdate)
group by r.name,
r.createdate           

其实这个问题是个递归问题，由上一个找到下一个，但是得构造一下：

declare @table1 table(name nvarchar,createdate smalldatetime)

insert into @table1
select ‘a‘,‘2011-03-01 10:00:00‘
union all select ‘a‘,‘2011-03-01 11:00:00‘
union all select ‘a‘,‘2011-03-01 12:00:00‘
union all select ‘a‘,‘2011-03-01 12:20:00‘

union all select ‘b‘,‘2011-03-01 13:10:00‘
union all select ‘b‘,‘2011-03-01 13:20:00‘
union all select ‘b‘,‘2011-03-01 14:30:00‘
union all select ‘b‘,‘2011-03-01 15:15:00‘
union all select ‘b‘,‘2011-03-01 16:00:00‘
union all select ‘b‘,‘2011-03-01 17:20:00‘
union all select ‘b‘,‘2011-03-01 17:15:00‘
union all select ‘b‘,‘2011-03-01 19:16:00‘
union all select ‘b‘,‘2011-03-01 17:15:00‘


;with a  --按照name分区，同时按照createdate排序编号
as
(
select name,
createdate,
ROW_NUMBER() over(partition by name
order by createdate) as k1
from @table1 
),

c   --对于每个时间,找到大于这个时间2小时的时间中最小那个时间
as
(
select a1.name,
a1.createdate,
a1.k1,
MIN(a2.createdate) as nextCreatedate,
MIN(a2.k1) as nextK1
from a a1
inner join a a2
on a1.name = a2.name
and a2.createdate >= DATEADD(hour,2,a1.createdate)

group by a1.name,
a1.createdate,
a1.k1

union all

select a.name,null,null,a.createdate,1  --构造递归运行时需要的层级
from a
where k1 = 1
),

w   --递归查询
as
(
select c.name,
c.createdate,
c.k1,
c.nextCreatedate,
c.nextK1,
1 as lev
from c
where createdate is null
and k1 is null

union all

select c.name,
c.createdate,
c.k1,
c.nextCreatedate,
c.nextK1,
lev + 1
from W
inner join c
on w.name = c.name
and w.nextCreatedate = c.createdate

)

SELECT distinct name,
nextCreatedate,
nextK1 
FROM w

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原文地址：https://www.cnblogs.com/lonelyxmas/p/12019942.html