最简单的方法是 lead() --假设你每天只有不超过一行，并且没有间隔 workday 列。所以，要获得这样的假期的第一天：

select er.*
from (select er.*,
             lead(workday, 4) over (partition by staffid order by workday) as workday_4
      from dbo.EmployeeRecords er
      where reasonId = 3 and isPaid = 1
     ) er
where workday_4 = dateadd(day, 4, workday);

“4”比“5”小一个，因为当前日期已计为一天。
如果你中间有周末，需要考虑这些，那么5天就是一周。您可以使用：

where workday_4 <= dateadd(day, 6, workday);

因为5天等于一周，所以这很容易奏效。否则你需要考虑到一周中的哪一天。

使用带有count运算符的子查询尝试下面的查询

select * from dbo.EmployeeRecords 
where staffid in 
(
    select staffid 
        from 
        (
        select staffid, count(staffid) cnt rownum
        from dbo.EmployeeRecords 
        where reasonId = 3 and isPaid = 1
        group by staffid
        )t
        where t.cnt>=5
)