（更新）按规律取数据？

[liu_ke_ya]

/*
有一张表（大约200万条记录）。为方便测试，剔除无关信息，随机生成10000行记录，保留3列，记为：
test（usrid，value，u_type）,其中usrid唯一，value在1000范围以内，u_type为‘Yes'或‘No’。

--要求：选择表中value值相同但u_type不同的记录，将其usrid成对找出来。
--例如：原始表为：
usrid    value    u_type
1          1        Yes 
2          34       No
4          86       No
5          34       No
6          7        Yes
8          1        Yes
9          1        No
3          10       Yes
89         10       Yes
78         7        No
14         2        No
66         2        Yes
102        2        No
708        8        Yes
84         8        No
99         8        Yes
182        8        No

则，最终表为（只有1行）：
Usrid    
1
9             
6
78 
14            
66            
84
99
182
708  

这里像value为1的记录，u_type有2个Yes，1个No。属于多对1，那么任意挑一个Yes和No的记录，找出其usrid（1和9）。
value为2的记录属于1对多,做类似处理。但是多对多的时候，要取Yes和No最小记录数，例如value为8时，
有2个Yes，2个No，那么都要取出,
即：对于每一个相同的value，取出的记录数是：2*min（Yes，No)。


*/

--随机生成数据
if OBJECT_ID('test2') is not null
drop table test2
go

create table test2(usrid int,value int,u_type varchar(5))

declare @i int
set @i=1

while @i<=10000
begin
    insert into test2 values(@i,ABS(CHECKSUM(newid())%1000),ABS(CHECKSUM(newid()))%2)
    set @i=@i+1
end

update test2
set u_type=case u_type when 1 then 'Yes'
               when 0 then 'No' end from test2 

--select count(1) from test2 

[问苍天人生几何]

;with cte as
(
	select value,sum(case when u_type='Yes' then 1 else 0 end) Yes,sum(case when u_type='No' then 1 else 0 end) No  from @test
	group by value
	having count(distinct u_type)>1
)
select t.usrid from
(
	select a.usrid,a.value,a.u_type,row_number() over(partition by a.value,a.u_type order by newid()) rn from @test a,cte b
	where a.value=b.value and b.Yes>b.No
) t
where t.rn=1
union all
select t.usrid from
(
	select a.usrid,a.value,a.u_type,row_number() over(partition by a.value,a.u_type order by a.usrid) rn from @test a,cte b
	where a.value=b.value and b.Yes<b.No
) t
where t.rn=1
union all
select t.usrid from
(
	select a.usrid,a.value,a.u_type,row_number() over(partition by a.value,a.u_type order by a.value) rn from @test a,cte b
	where a.value=b.value and b.Yes=b.No
) t
order by usrid

[liu_ke_ya]

[Quote=引用 1 楼 的回复:]

SQL code

;with cte as
(
select value,sum(case when u_type='Yes' then 1 else 0 end) Yes,sum(case when u_type='No' then 1 else 0 end) No from @test
group by value
having count(distinct u……
[/Quote]
当b.Yes>b.No或b.Yes<b.No时，应该去min（YES，no）条记录，比如Yes有3，No有2，那么应该取出2个Yes和2个No。您使用rn=1只取出了一个YES和一个NO？

[问苍天人生几何]

[Quote=引用 2 楼 的回复:]
引用 1 楼 的回复:

SQL code

;with cte as
(
select value,sum(case when u_type='Yes' then 1 else 0 end) Yes,sum(case when u_type='No' then 1 else 0 end) No from @test
group by value
having count(dist……
[/Quote]
哦，那我误解了。很简单，你将rn=1改成rn=b.No就可以了

[liu_ke_ya]

[Quote=引用 3 楼 的回复:]

引用 2 楼 的回复:
引用 1 楼 的回复:

SQL code

;with cte as
(
select value,sum(case when u_type='Yes' then 1 else 0 end) Yes,sum(case when u_type='No' then 1 else 0 end) No from @test
group by value
ha……
[/Quote]
谢谢提点，做了下更改，将三种情况合并了

	select 
		 value,
		 sum(case when u_type='Yes' then 1 else 0 end) Yes,
		 sum(case when u_type='No' then 1 else 0 end) No 
	into temp1
	from test2
	group by value
	having count(distinct u_type)>1
	
	select *
		,case when Yes>=No then No else Yes end as min_num
	into temp2
	from temp1
	
	select  t.* 
	from
	(
		select
		t1.usrid,
		t1.value,
		t1.u_type,
		min_num,
		row_number() over(partition by t1.value,t1.u_type order by t1.usrid) rin 
		from 
			test4 t1
		inner join 
			temp2 t2
		on t1.value=t2.value 
	) t
	where t.rin<=min_num