数据库 2009-6-9 0:02:47
2009-6-8 0:02:47
2009-6-4 0:02:47
2009-6-5 0:02:47
2009-6-9 0:02:47
2009-6-1 0:02:47
select url,title from News where intime like '%6-9%'没有查询结果
应该怎么写?
2009-6-8 0:02:47
2009-6-4 0:02:47
2009-6-5 0:02:47
2009-6-9 0:02:47
2009-6-1 0:02:47
select url,title from News where intime like '%6-9%'没有查询结果
应该怎么写?
select url,title from News where substring(convert(varchar(10),intime,20),6,5)='06-09'改成这样就可以了
insert @t select '2009-6-9 0:02:47'
insert @t select '2009-6-8 0:02:47'
insert @t select '2009-6-4 0:02:47'
insert @t select '2009-6-5 0:02:47'
insert @t select '2009-6-9 0:02:47'
insert @t select '2009-6-1 0:02:47'
select * from @t where convert(varchar(10),dt,120) like '%06-09%'
/*dt
------------------------------------------------------
2009-06-09 00:02:47.000
2009-06-09 00:02:47.000(影響 2 個資料列)
*/
比如你需要的
select url,title from News where to_char(intime ,'mm-dd') like '%6-9%'