Question

除了每个组合（channel / cdn / mediaplaylist）最近的2个记录之外，删除所有记录的SQL查询是什么。

例如，我有下表：

channel       cdn        mediaplaylist      response    date

channel1      cdn1       mediaplaylist1     12          2015-02-09 10:40:00
channel2      cdn2       mediaplaylist2     19          2015-02-09 10:41:00   
channel3      cdn3       mediaplaylist3     16          2015-02-09 10:42:00
channel1      cdn1       mediaplaylist1     20          2015-02-09 10:43:00
channel1      cdn1       mediaplaylist1     57          2015-02-09 10:46:00
channel2      cdn2       mediaplaylist2     83          2015-02-09 10:47:00    
channel3      cdn3       mediaplaylist3     37          2015-02-09 10:48:00
channel1      cdn1       mediaplaylist1     43          2015-02-09 10:49:00
channel1      cdn1       mediaplaylist1     33          2015-02-09 10:50:00
channel2      cdn2       mediaplaylist2     79          2015-02-09 10:51:00    
channel3      cdn3       mediaplaylist3     19          2015-02-09 10:52:00
channel1      cdn1       mediaplaylist1     13          2015-02-09 10:53:00

我想得到以下结果：

channel       cdn        mediaplaylist      response    date

channel1      cdn1       mediaplaylist1     12          2015-02-09 10:40:00
channel1      cdn1       mediaplaylist1     20          2015-02-09 10:43:00
channel2      cdn2       mediaplaylist2     19          2015-02-09 10:41:00
channel2      cdn2       mediaplaylist2     83          2015-02-09 10:47:00
channel3      cdn3       mediaplaylist3     16          2015-02-09 10:42:00
channel3      cdn3       mediaplaylist3     37          2015-02-09 10:48:00

Answer 1

使用SQL-Server很简单，只需使用CTE和排名函数ROW_NUMBER：

WITH CTE AS
(
    SELECT RN = ROW_NUMBER() OVER (Partition By channel, cdn, mediaplaylist 
                                Order By date DESC), 
           c.*  -- select all columns for debugging purposes
    FROM dbo.Channels c
)
DELETE FROM CTE WHERE RN > 2

我喜欢common-table-expressions，因为它们允许看到我要删除的内容。因此，您只需将DELETE替换为SELECT *。

Answer 2

以下查询按您的方式枚举行：

select f.*,
       (@rn := if(@ccp = concat_ws(':', channel, cdn, mediaplaylist), @rn + 1,
                  if(@ccp := concat_ws(':', channel, cdn, mediaplaylist), 1, 1)
                 )
       ) as seqnum
from following f cross join
     (select @rn := 0, @ccp := '') vars
order by channel, cdn, mediaplaylist, date desc;

您希望seqnum为1或2的值。

您可以将其放在delete：

中

删除f 从以下f加入（选择f。*，（@rn：= if（@ccp = concat_ws（'：'，channel，cdn，mediaplaylist），@ rn + 1， if（@ccp：= concat_ws（'：'，channel，cdn，mediaplaylist），1,1）））作为seqnum 从以下f交叉加入（选择@rn：= 0，@ ccp：=''）vars 按渠道，cdn，媒体播放列表，日期desc ）ccp 在f.channel = ccp.channel和f.cdn = ccp.cdn和f.mediaplaylist = cdn.mediaplaylist 其中seqnum＆lt; 3;

sql：删除除最近2个以外的所有记录（channel / cdn / mediaplaylist）

2 个答案: