日期:2014-05-17  浏览次数:20392 次

(更新)按规律取数据?
SQL code

/*
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 




------解决方案--------------------
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_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

------解决方案--------------------
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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……