如果它是一首又一首歌曲,假设一个名为tblSongs并带有'序列'&'名称'列的表。你可能想尝试像
select top N first.name, second.name, count(*)
from tblSongs as first
inner join tblSongs as second
on second.sequence=first.sequence + 1
group by first.name, second.name
order by count(*) desc
如果歌曲序列X,Y被计一样Y,X,则
select top N first.name, second.name, count(*)
from tblSongs as first
inner join tblSongs as second
on second.sequence=first.sequence + 1
or second.sequence=first.sequence - 1
group by first.name, second.name
order by count(*) desc
如果您正在寻找2个歌曲序列的任何模式,那么
select first.name, second.name, abs(second.sequence - first.sequence) as spacing_count
from tblSongs as first
inner join tblSongs as second
on second.sequence=first.sequence + 1
or second.sequence=first.sequence - 1
然后做一些关于spacing_count(这超出了我)的统计分析。
我相信那些会让你开始。
谢谢,将尽快尝试! – staqUUR 2009-08-12 01:29:49
那么,在一个洞!谢谢约翰,我相信你给了我想要做的事情的基础。现在进行费力的手动交叉检查(叹气) – staqUUR 2009-08-12 04:06:00