的Oracle SQL条件选择查询

Question

我有此表下面

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) 
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) 

我可以用一个正常的选择查询来提取数据，使得它成为

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) 

的date_time的字段以递增的顺序。逻辑是time_type'1'需要与同一个rid的下一个time_type'2'配对。如果time_type'1'或'2'出现在由date_time排序的2个或更多的组中，我将采用较早的一个并忽略其余部分。

可以这样做吗？

Answer 1

尝试此查询：

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; 

这里是一个SQL Fiddle demo