Question

我建立了一个CASE WHEN表达式，以显示某人在其用户ID下系统中没有系统事务超过15分钟的情况。

  case
    when MOD_DATE_TIME > prior MOD_DATE_TIME+(1/96) 
         and user_id = prior user_id 
    then round((MOD_DATE_TIME - prior MOD_DATE_TIME)*1440,2)
    else null
  end as TIME_GAP

以上是仅声明时的情况。完整的查询如下：

 select
  user_id, MENU_OPTN_NAME, MOD_DATE_TIME,
  case
    when MOD_DATE_TIME > prior MOD_DATE_TIME+(1/96) and user_id = prior user_id then round((MOD_DATE_TIME - prior MOD_DATE_TIME)*1440,2)
    else null
  end as TIME_GAP
from
  (select
     ptt.user_id, MENU_OPTN_NAME, ptt.MOD_DATE_TIME,
     row_number() over (partition by ptt.user_id order by ptt.MOD_DATE_TIME) seq
   from PROD_TRKG_TRAN ptt
     join cd_master cm on
       ptt.cd_master_id = cm.cd_master_id
   Where
     MENU_OPTN_NAME = 'Cycle Cnt {Reserve}' --CHANGE BASED ON WHAT YOUR TRACKING... MENU NAMES AT THE BOTTOM
     and ptt.user_id = 'LLEE1' --CHANGE BASED ON WHO YOU WANT TO TRACK FOR A GAP...
     and cm.cd_master_id =
       (select cd_master_id
        from cd_master
        where
          co = '&CO'
          and div = '&DIV')
     and ptt.create_date_time >=
/*Today*/ trunc(sysdate)
--/*This Week*/              trunc(sysdate-(to_char(sysdate,'D')-1))
     --/*This Month*/            trunc(sysdate)-(to_char(sysdate,'DD')-1)
     --/*Date Range*/            '&FromDate' and ptt.create_date_time-1 < '&ToDate'
     --group by ptt.user_id
)cc
CONNECT BY
  user_id = prior user_id
  and seq = prior seq+1
start with
  seq = 1

我要查询执行的操作在CASE WHEN子句中。我希望它能将轮班开始时间定为下午4点，因此，如果他们在下午4:41之前不做任何事情，那么我会认为这是41分钟的间隔。

在Oracle中的情况

0 个答案: