我如何在sql中计算封闭季度

如果 SQL Server 的版本是 2012 或更高版本，则可以使用LAG功能。

select
  s.[date], s.fio, s.salary,
  datepart(quarter, s.[date]) quarter_nbr,
  j1.quarter_salary
from salaries s
join (
  select
    a2.prev_quarter, a1.cur_quarter, a2.next_quarter,
    case count(distinct a1.cur_month)
      when 3 then sum(s.salary)
      else case
             when lag(
                    count(distinct a1.cur_month)
                  ) over(order by a1.cur_quarter) = 3 and
                  lag(
                    a1.cur_quarter
                  ) over(order by a1.cur_quarter) = a2.prev_quarter
             then lag(sum(s.salary)) over(order by a1.cur_quarter)
             else 0
           end
    end quarter_salary
  from salaries s cross apply (
    select dateadd(quarter, datediff(quarter, 0, [date]), 0) cur_quarter,
           dateadd(month, datediff(month, 0, [date]), 0) cur_month
  ) a1 cross apply(
    select dateadd(quarter, -1, cur_quarter) prev_quarter,
           dateadd(quarter, 1, cur_quarter) next_quarter
  ) a2
  group by a2.prev_quarter, a1.cur_quarter, a2.next_quarter
) j1
on s.[date] >= j1.cur_quarter and s.[date] < j1.next_quarter;

结论：

+------------+---------+--------+-------------+----------------+
|    date    |   fio   | salary | quarter_nbr | quarter_salary |
+------------+---------+--------+-------------+----------------+
| 2018-12-31 | Сидоров |   2000 |           4 |           4400 |
| 2018-11-28 | Иванов  |   1200 |           4 |           4400 |
| 2018-10-31 | Иванов  |   1200 |           4 |           4400 |
| 2019-06-30 | Иванов  |   1200 |           2 |              0 |
| 2019-09-30 | Сидоров |   1400 |           3 |           9600 |
| 2019-09-16 | Иванов  |   1800 |           3 |           9600 |
| 2019-08-31 | Сидоров |   2000 |           3 |           9600 |
| 2019-08-31 | Иванов  |   1200 |           3 |           9600 |
| 2019-07-31 | Сидоров |   2000 |           3 |           9600 |
| 2019-07-31 | Иванов  |   1200 |           3 |           9600 |
| 2019-10-14 | Сидоров |   2900 |           4 |           9600 |
| 2019-10-14 | Иванов  |   1800 |           4 |           9600 |
+------------+---------+--------+-------------+----------------+

SQL Fiddle上的演示。