日期:2014-05-17 浏览次数:20484 次
/* 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。如果有1对多,多对1, -- 多对多情形,则任选其一,使其成为1对1. --例如:如果原始表为: 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 最终表为(2行): Usrid_Yes Usrid_No 1 9 6 78 66 14 99 84 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2,8的记录属于1对多,多对多,做类似处理。 --说明一下,使用类似下面的cross join的方法,大数据量时,不太可行。 select t1.usr , t1.u_type, t2.usr , t2.u_type from test2 t1,test2 t2 where t1.value=t2.value and t1.u_type!=t2.u_type and t1.usr!=t2.usr order by t1.usr,t2.usr 。 。 。 */ --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
-->try select a.usrid,a.value from (select row_number() over(partition by value order by value) rn,* from test2)a inner join (select value from test2 group by value having count(distinct u_type)>1) b on a.value=b.value where a.rn=1
------解决方案--------------------
with tt as (
select value
from test
group by value
having COUNT(distinct u_type)>1
)
select
(select top 1 usrid from test t where t.value=tt.value and t.u_type='Yes') Usrid_Yes ,
(select top 1 usrid from test t where t.value=tt.value and t.u_type='No') Usrid_No
from tt
------解决方案--------------------
with t as ( select value,u_type,usrid, ROW_NUMBER() over (partition by value,u_type order by usrid) as rn from test2 ),tYes as( select * From t where u_type='Yes' and rn=1 ) ,tNo as ( select * From t where u_type='No' and rn=1 ) select a.usrid as Usrid_Yes,b.usrid as Usrid_No from tYes a join tNo b on a.value=b.value order by a.usrid