Question

我要在行中获取所有输入/输出事件。我的表如下所示；

Trnevents：

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document.getElementById("submit").onclick = function () { myFunction() };
function myFunction() {
    $.get('@Url.Action("FromInput", "Home")', function(data) {
    $('#output').replaceWith(data);
    });
}

我在下面的查询中尝试过，但对该输出不满意

 emp_id EVENTID     DT
 4176   3   2019-04-09 15:34:12.000
 4176   1   2019-04-09 08:55:13.000
 4176   3   2019-04-08 16:08:32.000
 4176   2   2019-04-08 15:08:18.000
 4176   1   2019-04-08 14:41:37.000
 4176   0   2019-04-08 08:45:13.000

预期像：

 select
t.emp_reader_id as empId,B.emp_name, cast(max(DT) as date) as Belongs_to,B.areaname as POINTID,
 max(iif(EVENTID = 0, DT, null)) as F1, max(iif(EVENTID = 1, DT, null)) as F2, max(iif(EVENTID = 2, DT, null)) as F3, max(iif(EVENTID = 3, DT, null)) as F4,
  format(dateadd(ss,Datediff(second,max(iif(EVENTID = 0, DT, null)),max(iif(EVENTID = 3, DT, null))),0),'HH:mm') as Worked from (   
select
    *, grp = sum(iif(EVENTID = 0, 3, 0) ) over (partition by emp_reader_id order by DT)
from        Trnevents ) t inner join employee B on t.emp_reader_id=B.emp_reader_id where b.emp_reader_id=4176
   group by t.emp_reader_id,t.grp,B.emp_name,t.Belongs_to,B.areaname
                  order by emp_reader_id,t.Belongs_to,F1,F2 asc

我需要如上所述的输出。谢谢！

Eventid：

0作为F1， 1作为F2， 2作为F3， 3作为F4

Answer 1

下面的代码段应该有帮助：

SELECT emp_id,day,
TO_CHaR(to_date(F1,'HH24:MI'),'HH24:MI') F1,
TO_CHaR(to_date(F2,'HH24:MI'),'HH24:MI') F2,
TO_CHaR(to_date(F3,'HH24:MI'),'HH24:MI') F3,
TO_CHaR(to_date(F4,'HH24:MI'),'HH24:MI') F4  FROM (
  SELECT emp_id,event_id,to_char(DT,'DD-MON-YYYY') day ,to_number(replace(to_char(DT,'HH24:MI'),':','.'),'99.99') dt_time from sample_event a

)
PIVOT (
  avg(dt_time)
  FOR event_id in (
    0 F1,1 F2,2 F3,3 F4  )
)

ORDER BY DAY asc;

-- output
EMP_ID  DAY         F1      F2      F3      F4
4176    08-APR-2019 08:45   14:41   15:08   16:08
4176    09-APR-2019  -      08:55    -      15:34

Answer 2

以下内容是在oracle数据库上。因此，不要对“ from dual”感到困惑，而以开头只是简单地模拟输入数据。同样在oracle trunc（date_time）上返回仅日期部分。首先创建mydata的是模拟输入。接下来，通过按日期获取绝对最小值/最大值获取f1f4。 Next通过说我们只想要日期与f1和f4不匹配的那些行来计算f2和f4。根据您的规则，emp + dat有2或4行。然后，final将原始源与这2个源连接起来，得到f1，f2，f3，f4。区别在于驱动程序是原始的

with mydata as (
    select '4176' emp_id, '3' eventid, to_date('2019-04-09 15:34:12', 'YYYY-MM-DD HH24:MI:SS') dt from dual union all
    select '4176' emp_id, '1' eventid, to_date('2019-04-09 08:55:13', 'YYYY-MM-DD HH24:MI:SS') dt from dual union all
    select '4176' emp_id, '3' eventid, to_date('2019-04-08 14:41:37', 'YYYY-MM-DD HH24:MI:SS') dt from dual union all
    select '4176' emp_id, '2' eventid, to_date('2019-04-08 15:08:18', 'YYYY-MM-DD HH24:MI:SS') dt from dual union all
    select '4176' emp_id, '1' eventid, to_date('2019-04-08 14:41:37', 'YYYY-MM-DD HH24:MI:SS') dt from dual union all
    select '4176' emp_id, '0' eventid, to_date('2019-04-08 08:45:13', 'YYYY-MM-DD HH24:MI:SS') dt from dual
),
getf1f4 as (
    select emp_id, trunc(dt) dt, count(*) num_rows, min(dt) f1, max(dt) f4
    from mydata
    group by emp_id, trunc(dt)
),
getf2f3 as (
    select ee.emp_id, getf1f4.dt,
           min(ee.dt) f2,
           max(ee.dt) f3
    from mydata ee join getf1f4 on ee.emp_id = getf1f4.emp_id and trunc(ee.dt) = getf1f4.dt
    where ee.dt not in ( getf1f4.f1, getf1f4.f4)
    group by ee.emp_id, getf1f4.dt
)
select distinct 
       ee.emp_id, getf1f4.dt, getf1f4.f1, getf2f3.f2, getf2f3.f3, getf1f4.f4
from mydata ee 
     join getf1f4 on ee.emp_id = getf1f4.emp_id and trunc(ee.dt) = getf1f4.dt
     left outer join getf2f3 on ee.emp_id = getf2f3.emp_id and trunc(ee.dt) = getf2f3.dt

Answer 3

@dolu bolu

可以从以下查询中引用和更改所请求的sql查询：

select emp_id, day, f1, f2, f3, f4  from 
(select a.emp_id,to_char(a.DT,'DD-MON-YYYY') day,
(select to_char(b.DT,'HH24:MI') from sample_event b where b.emp_id=a.emp_id and to_char(b.DT,'DD-MON-YYYY')=to_char(a.DT,'DD-MON-YYYY') and b.event_id=0 ) F1,
(select to_char(b.DT,'HH24:MI') from sample_event b where b.emp_id=a.emp_id and to_char(b.DT,'DD-MON-YYYY')=to_char(a.DT,'DD-MON-YYYY') and b.event_id=1 ) F2,
(select to_char(b.DT,'HH24:MI') from sample_event b where b.emp_id=a.emp_id and to_char(b.DT,'DD-MON-YYYY')=to_char(a.DT,'DD-MON-YYYY') and b.event_id=2 ) F3,
(select to_char(b.DT,'HH24:MI') from sample_event b where b.emp_id=a.emp_id and to_char(b.DT,'DD-MON-YYYY')=to_char(a.DT,'DD-MON-YYYY') and b.event_id=3 ) F4,
(select max(b.DT) from sample_event b where b.emp_id=a.emp_id and to_char(b.DT,'DD-MON-YYYY')=to_char(a.DT,'DD-MON-YYYY')  ) max_time,
(select min(b.DT) from sample_event b where b.emp_id=a.emp_id and to_char(b.DT,'DD-MON-YYYY')=to_char(a.DT,'DD-MON-YYYY')  ) min_time
from sample_event a group by a.emp_id,to_char(a.DT,'DD-MON-YYYY'))
order by day;

您可以使用datediff函数获取max_time和min_time之间的小时数

如何按日期分组并获得全部进出时间？

3 个答案: