日期:2014-05-18 浏览次数:20535 次
DECLARE @T table ( code varchar(20), ID INT ) INSERT INTO @t SELECT 'AAA',1 UNION ALL SELECT 'AAA',3 UNION ALL SELECT 'AAA',4 UNION ALL SELECT 'BBB',2 UNION ALL SELECT 'BBB',3 SELECT *, P_ID = (SELECT MAX(ID) FROM @t WHERE code = A.code AND ID<A.ID), N_ID = (SELECT MIN(ID) FROM @t WHERE code = A.code AND ID>A.ID) FROM @t A
------解决方案--------------------
create table tb(CODE varchar(10),ID int) insert into tb values('AAA', 1 ) insert into tb values('AAA', 3 ) insert into tb values('AAA', 4 ) insert into tb values('BBB', 2 ) insert into tb values('BBB', 3 ) go select m.code,m.id,n.previous_id,m.next_id from ( select t1.*,t2.id next_ID from (select px=(select count(1) from tb where CODE=a.CODE and id<a.id)+1 , * from tb a) t1 left join (select px=(select count(1) from tb where CODE=a.CODE and id<a.id)+1 , * from tb a) t2 on t1.code = t2.code and t1.px = t2.px - 1 ) m, ( select t1.px,t1.code,t2.id previous_ID from (select px=(select count(1) from tb where CODE=a.CODE and id<a.id)+1 , * from tb a) t1 left join (select px=(select count(1) from tb where CODE=a.CODE and id<a.id)+1 , * from tb a) t2 on t1.code = t2.code and t1.px - 1 = t2.px ) n where m.code = n.code and m.px = n.px drop table tb /* code id previous_id next_id ---------- ----------- ----------- ----------- AAA 1 NULL 3 AAA 3 1 4 AAA 4 3 NULL BBB 2 NULL 3 BBB 3 2 NULL (所影响的行数为 5 行) */
------解决方案--------------------
DECLARE @T table ( code varchar(20), ID INT ) INSERT INTO @t SELECT 'AAA',1 UNION ALL SELECT 'AAA',3 UNION ALL SELECT 'AAA',4 UNION ALL SELECT 'BBB',2 UNION ALL SELECT 'BBB',3 select *, nId=identity(int,1,1) into #T from @t order by code,ID select T.code, T.ID, Tper.ID, Tnex.ID from #T T left join #T Tper on T.nId=Tper.nId+1 and T.code=Tper.code left join #T Tnex on T.nId=Tnex.nId-1 and T.code=Tnex.code drop table #T /* code ID ID ID -------------------- ----------- ----------- ----------- AAA 1 NULL 3 AAA 3 1 4 AAA 4 3 NULL BBB 2 NULL 3 BBB 3 2 NULL (5 row(s) affected) */
------解决方案--------------------
create table tb(CODE varchar(10),ID int) insert into tb values('AAA', 1 ) insert into tb values('AAA', 3 ) insert into tb values('AAA', 4 ) insert into tb values('BBB', 2 ) insert into tb values('BBB', 3 ) 借用临时表方法: select * into #temp from (select bh=(select count(1) from tb where code=a.code and id<a.id)+1,* from tb a)b select code,id,previous_ID=(select id from #temp where code=a.code and bh=a.bh-1), next_ID=(select id from #temp where code=a.code and bh=a.bh+1) from #temp a
------解决方案--------------------