Oracle SQL条件选择查询
我有这张桌子在下面Oracle SQL条件选择查询,sql,oracle,select,plsql,Sql,Oracle,Select,Plsql,我有这张桌子在下面 uid rid time_type date_time a11 1 2 5/4/2013 00:32:00 (row1) a43 1 1 5/4/2013 00:32:01 (row2) a68 1 1 5/4/2013 00:32:02 (row3) a98 1 2 5/4/2013 00:32:03 (row4)
uid rid time_type date_time
a11 1 2 5/4/2013 00:32:00 (row1)
a43 1 1 5/4/2013 00:32:01 (row2)
a68 1 1 5/4/2013 00:32:02 (row3)
a98 1 2 5/4/2013 00:32:03 (row4)
a45 1 2 5/4/2013 00:32:04 (row5)
a94 1 1 5/4/2013 00:32:05 (row6)
a35 1 2 5/4/2013 00:32:07 (row7)
a33 1 2 5/4/2013 00:32:08 (row8)
我是否可以使用普通的select查询来提取数据,使其成为
uid rid time_type date_time
a43 1 1 5/4/2013 00:32:01 (row2)
a98 1 2 5/4/2013 00:32:03 (row4)
a94 1 1 5/4/2013 00:32:05 (row6)
a35 1 2 5/4/2013 00:32:07 (row7)
日期\时间字段按asc顺序排列。逻辑是时间类型“1”需要与同一rid的下一时间类型“2”配对。如果时间类型“1”或“2”出现在按日期时间排序的2个或更多的组中,我将选择较早的一个,忽略其余的
可以这样做吗?尝试以下查询:
with src as (
select tst.*,
case when time_type <> lag( time_type) over ( partition by rid order by date_time, time_type )
then 1 else 0
end take_me
from tst
)
select * from src where take_me = 1
order by rid, date_time;
这是一个标签为mysql和Oracle的。你用哪一种?