use [tongxing]
declare @days int
set @days=1000 --表示跟今天相差的天數,今天為1,乙個星期內為7..
select (case when iclassid=1 then '新聞' when iclassid=2 then '生活' when iclassid=3 then '軍事' when iclassid=4 then '科技' when iclassid=5 then '體育' when iclassid=6 then '財經' when iclassid=7 then '娛樂' when iclassid=8 then '文學' else '遊戲' end) as 欄目,count(*) as 數量
from [digg_article] where datediff(day,daddtime,getdate())<@days
group by iclassid
select @days as 相差天數
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