我在下面有这张表
uid rid time_type date_time
a11 1 1 5/4/2013 00:32:00 (row1)
a43 2 1 5/4/2013 00:32:01 (row2)
a68 2 2 5/4/2013 00:32:02 (row3)
a98 2 1 5/4/2013 00:32:03 (row4)
a45 2 1 5/4/2013 00:32:04 (row5)
a94 1 2 5/4/2013 00:32:05 (row6)
a35 2 2 5/4/2013 00:32:07 (row7)
a33 2 2 5/4/2013 00:32:08 (row8)
我可以使用普通的选择查询来提取数据,使其变为
uid rid time_type date_time
a11 1 1 5/4/2013 00:32:00 (row1)
a94 1 2 5/4/2013 00:32:05 (row6)
a43 2 1 5/4/2013 00:32:01 (row2)
a68 2 2 5/4/2013 00:32:02 (row3)
a98 2 1 5/4/2013 00:32:03 (row4)
a35 2 2 5/4/2013 00:32:07 (row7)
a45 2 1 5/4/2013 00:32:04 (row5)
a33 2 2 5/4/2013 00:32:08 (row8)
逻辑是 1 的 time_type 需要与下一个对应的 time_type 2 配对才能实现相同的摆脱。这可以做到吗?