如上图所示:
1、错位相减:time2的时间减去time1的时间,比如“2012-03-19 10:18:53” - “2012-03-19 10:09:33”
2、排序方式是 time1 desc
3、time1 desc 排序后,ID与A的内容不一定是按照顺序下来的,但一定是从大到小排下来的。
比较啰嗦
1、错位相减:time2的时间减去time1的时间,比如“2012-03-19 10:18:53” - “2012-03-19 10:09:33”
2、排序方式是 time1 desc
3、time1 desc 排序后,ID与A的内容不一定是按照顺序下来的,但一定是从大到小排下来的。
比较啰嗦
解决方案 »
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ID A Time1 Time2
12 D203456700 2012-03-19 10:08:01 2012-03-19 10:10:01
11 D203456600 2012-03-19 10:06:01 2012-03-19 10:08:01
9 D203456500 2012-03-19 10:04:01 2012-03-19 10:06:01
8 D203456300 2012-03-19 10:02:01 2012-03-19 10:04:01
” - “2012-03-19 10:06:01 ”
用SQL语句怎样写?
time2-time1
varchar
to_date(time2,'YYYY-MM-DD hh24:mi:ss')-to_date(time1,'YYYY-MM-DD hh24:mi:ss')
得到的结果的单位是天……
你误解了我的意思,
比如:
ID A time1 time2
3 D202345600 10(时间格式) 11
2 D202345500 9 10
我想要的结果是:time2-time1,即11-9,每次都这样计算,相减后作为一列值。
SQL语句写法?
select 12 ID, 'D203456700' A, '2012-03-19 10:08:01' Time1,'2012-03-19 10:10:01' Time2 from dual union all
select 11, 'D203456600' ,'2012-03-19 10:06:01' ,'2012-03-19 10:08:01' from dual union all
select 9 ,'D203456500' ,'2012-03-19 10:04:01' ,'2012-03-19 10:06:01' from dual union all
select 8 ,'D203456300' ,'2012-03-19 10:02:01' ,'2012-03-19 10:04:01' from dual)select prior to_date(time2,'yyyy-mm-dd hh24:mi:ss')- to_date(time1,'yyyy-mm-dd hh24:mi:ss')
from tb start with Time2='2012-03-19 10:10:01' connect by Time2=prior Time1
select b.time2 begintime, a.time1 endtime, (b.time2 - a.time1)*24*60||' minutes' res
from
(select id, a, time1, time2 , rownum as cnt from test_yixl order by time1 desc) a,
(select id, a, time1, time2 , rownum as cnt from test_yixl order by time1 desc) b
where b.cnt + 1 = a.cnt;