日期:2014-05-18 浏览次数:20485 次
select x.*,case when x.sumhit>y.sumhit then '上升' else '下降' end 相对昨日 from ( select a.articleid,posteddate,mh maxhit,sh sumhit from tb a inner join ( select id,max(hit) mh,sum(hit) sh from tb group by articleid ) b on a.id=b.id ) x left join ( select a.articleid,posteddate,mh maxhit,sh sumhit from tb a inner join ( select id,max(hit) mh,sum(hit) sh from tb group by articleid ) b on a.id=b.id ) y on datediff(dd,y.posteddate,x.posteddate)=1
------解决方案--------------------
create table table1(id int,articleid int,hit int, caption varchar(10),posteddate datetime)
insert into table1 select 1 , 10 , 1 , 'a' , '2007-10-22'
insert into table1 select 2 , 10 , 10 , 'a' , '2007-10-23'
insert into table1 select 3 , 9 , 5 , 'b' , '2007-10-22'
insert into table1 select 4 , 9 , 4 , 'b' , '2007-10-23'
insert into table1 select 5 , 8 , 4 , 'c' , '2007-10-23'
insert into table1 select 6 , 7 , 4 , 'd' , '2007-10-22'
select a.articleid,a.hit [hit(今日点击)],(select max(hit) from table1 where caption=a.caption) [maxhit(最大点击)],
convert(varchar(10),a.posteddate,120) [maxHitDate(最大点击日期)],(select sum(hit) from table1 where caption=a.caption)[SumHit(该id点击总数)],
case when a.hit-isnull((select hit from table1 where datediff(dd,posteddate,a.posteddate)=1 and caption=a.caption),0)>0 then '上升'
when a.hit-isnull((select hit from table1 where datediff(dd,posteddate,a.posteddate)=1 and caption=a.caption),0) <0 then '下降' else '平' end [相对昨日]
from table1 a,(select max(posteddate) posteddate,caption from table1 group by caption) b
where a.posteddate=b.posteddate and a.caption =b.caption order by articleid desc