SQL查詢可根據該范圍的行之前和之后的記錄轉換記錄集或記錄范圍

Question

假設這張桌子

Day Present Absent  Holiday
1/1/2019    1   0   0
1/2/2019    0   1   0
1/3/2019    0   0   1
1/4/2019    0   0   1
1/5/2019    0   0   1
1/6/2019    0   1   0
1/7/2019    1   0   0
1/8/2019    0   1   0
1/9/2019    0   0   1
1/10/2019   0   1   0

我想將缺席之間的所有假期標記為零，如果員工在假期之前和之后缺席，那么假期對他來說將成為缺席天數。 我不想使用循環，我想設置基本查詢方法。

Answer 1

作為select ，您可以使用lead()和lag() ：

select t.*,
       (case when prev_absent = 0 and next_absent = 0 and holiday = 1
             then 0 else holiday
        end) as new_holiday
from (select t.*,
             lag(absent) over (order by day) as prev_absent,
             lead(absent) over (order by day) as next_absent
      from t
     ) t;

如果這樣做符合您的要求，則可以將其合並到update ：

with toupdate as (
      select t.*,
             (case when prev_absent = 0 and next_absent = 0 and holiday = 1
                   then 0 else holiday
              end) as new_holiday
      from (select t.*,
                   lag(absent) over (order by day) as prev_absent,
                   lead(absent) over (order by day) as next_absent
            from t
           ) t
     ) t
update toupdate
    set holiday = new_holiday
    where holiday <> new_holiday;

編輯：

您也可以使用join s來做到這一點：

select t.*,
       (case when tprev.absent = 0 and tnext.absent = 0 and t.holiday = 1
             then 0 else holiday
        end) as new_holiday
from t left join
     t tprev
     on tnext.day = dateadd(day, -1, t.day) left join
     t tnext
     on tprev.day = dateadd(day, 1, tprev.day)