“像这样的东西”。嗯，最简单的方法是每十分钟从“日历时钟”中取出第一张唱片。这将是00:00、00:10、00:20等之后的第一个记录。
为此，您可以使用 first_value() 以及 select distinct :

select distinct
       first_value(x) over (partition by dd, hh, mm order by timestamp) as x,
       first_value(y) over (partition by dd, hh, mm order by timestamp) as y,
       first_value(receiver) over (partition by dd, hh, mm order by timestamp) as receiver,
       min(timestamp) over (partition by dd, hh, mm),
       count(*) over (partition by dd, hh, mm) as cnt

from t cross apply
     (values (convert(date, timestamp), datepart(hour, timestamp), datepart(minute, timestamp) / 6)
     ) v(dd, hh, mm);

这并不完全返回您指定的结果，但它似乎满足问题中提到的条件。
编辑：
我突然想到，从第一个时间戳开始，你可能需要10分钟的间隔。如果是：

select (case when seqnum = 1 then x end) as x,
       (case when seqnum = 1 then y end) as y,
       (case when seqnum = 1 then receiver end) as receiver,
       min(timestamp),
       count(*)
from (select t.*,
             datediff(second, min_ts, timestamp) / (10 * 60) as grp,
             row_number() over (partition by datediff(second, min_ts, timestamp) / (10 * 60) order by timestamp) as seqnum,
      from (select t.*,
                   min(timestamp) over () as min_ts
            from t
           ) t
     ) t
group by grp;

我认为这会返回您指定的组。按grp分组；

使用下面的查询，
结果1：

select t1.* from table t1
inner join t2
on (datediff(minute, starttime, endtime) = 10);

结果2：

select x, t, reciever, count(1)
from
(select t1.* from table t1
inner join t2
on (datediff(minute, starttime, endtime) = 10)) qry
group by x, t, reciever;