可以使用count和min窗口函数，如下所示：

with t as 
(
  select tb.*,
      count(*)  over  (partition by date_, code, Duration) cnt,
      min(Duration) over (partition by date_, code) min_dur
  from tbl tb
) 
select t.date_, t.code, t.duration,
  case 
    when min_dur = 0 or -- there is a 0 duration
         max(cnt) over (partition by date_, code) > 1 -- when there are the same durations in multiple rows. 
    then 'Y' else 'N'
  end as flg
from t

demo

我有点累了，所以可能有一个更简单的方法，我现在看不到，但你可以交叉SH和非SH的日期/持续时间，然后添加持续时间为零的日期。

select date, code, duration
from mytable
where date in
(
  select date from
  (
    select date, duration
    from mytable
    where code = 'SH'
    intersect
    select date, duration
    from mytable
    where code <> 'SH'
    union all
    select date, duration
    from mytable
    where duration = 0
  )
);

或

select date, code, duration
from mytable t
where exists
(
  select date, duration
  from mytable t2
  where t2.date = t.date and t2.code = 'SH'
  intersect
  select date, duration
  from mytable t3
  where t3.date = t.date and t3.code <> 'SH'
)
or date in
(
  select date
  from mytable
  where duration = 0
);