Question

我有一张表作为测试

with
     job_events ( userid, eventstartdt, jobcode ) as (
       select 10, date '2014-08-18', 'j1' from dual union all
       select 10, date '2015-07-27', 'j1' from dual union all
       select 10, date '2015-03-23', 'j2' from dual union all
       select 10, date '2015-12-28', 'j3' from dual union all
       select 10, date '2015-03-23', 'j4' from dual union all
       select 10, date '2015-03-23', 'j5' from dual union all
       select 35, date '2014-04-18', 'j1' from dual union all
       select 35, date '2015-09-22', 'j1' from dual union all
       select 35, date '2015-10-29', 'j8' from dual
     ),
     comp_events ( userid, eventstartdt, salary ) as (
       select 10, date '2014-08-11', 1000 from dual union all
       select 10, date '2015-03-23', 1525 from dual union all
       select 10, date '2015-06-21',  500 from dual union all
       select 10, date '2016-03-21', 2000 from dual union all
       select 35, date '2015-06-15', 2850 from dual
     ),
     u ( userid, eventstartdt, jobcode, salary ) as (
       select userid, eventstartdt, jobcode, null    from   job_events
       union all
       select userid, eventstartdt, null   , salary  from   comp_events
     ),
     prep ( userid, eventstartdt, jobcode, salary ) as (
       select userid, eventstartdt, jobcode, 
              last_value(salary ignore nulls) over 
                          (partition by userid order by eventstartdt)
       from   u
     )
select   userid, eventstartdt, jobcode, salary
from     prep
where    jobcode is not null
order by userid, jobcode, eventstartdt
;

Shiftend的数据类型为time out的数据类型为smalldatetime

我期待输出为

    USERID EVENTSTARTDT JOBCODE     SALARY
---------- ------------ ------- ----------
        10 2014-08-18   j1            1000
        10 2015-07-27   j1             500
        10 2015-03-23   j2            1525
        10 2015-12-28   j3             500
        10 2015-03-23   j4            1525
        10 2015-03-23   j5            1525
        35 2014-04-18   j1
        35 2015-09-22   j1            2850
        35 2015-10-29   j8            2850

9 rows selected.

我正在尝试此查询：

shiftend            |        out            |       
---------------------------------------------   
15:00:00.0000000    |   2016-07-22 14:42:00 |   
16:00:00.0000000    |   2016-07-22 16:06:00 |

但我得到的输出为

shiftend            |         out           |   Output    
-----------------------------------------------------------------------------   
15:00:00.0000000    |   2016-07-22 14:42:00 |   -00:18:00   
16:00:00.0000000    |   2016-07-22 16:06:00 |    00:06:00

23:42:00不正确。如何计算时间。

Answer 1

尝试以下查询：

select 
 case when (cast(out as time) < shiftend) then '-' else '' end + 
 convert(varchar(8), 
  dateadd(minute, 
   abs(
    DATEDIFF(minute, 
     cast(out as time)
    , shiftend)
   )
 ,0)
,108) as Output

说明：

您可以通过DATEDIFF(minute, cast(out as time), shiftend)获得两个日期之间的差异。
您只需要time组件即可避免前一天，因此您使用cast(out as time)。您提到的shiftend已经是数据类型time
abs返回绝对值，因此-18变为18。
然后使用00:00:00

dateadd(minute, [above value], 0)

最后convert(varchar(8),____,108)是因为您需要输出作为时间。
iif(cast(out as time) < shiftend,'-','')会在字词的开头添加否定号。

不幸的是，time数据类型中不能包含负值。它正在转变为-18至00：00，即23:42。你可以：

使用datediff并将差异保存为秒数，例如。
仅保存模块化差异，并有一个单独的列来确定它是正面还是负面的差异。
编写您自己的SQL函数，将其计算为varchar，如上所述

时间上的差异给出了不正确的结果

1 个答案: