尝试此查询。

SELECT id, Price, sum(Qt) FROM ticket
GROUP BY id,Price

你的产出；

id       Price       sum(Qt)
1        3.00        4
2        8.00        7

你可以用一个 group_concat() / substring_index() 技巧：

SELECT id, Price, SUM(Qt)
       SUBSTRING_INDEX(GROUP_CONCAT(price ORDER BY dtsell DESC), ',' 1) as last_price
FROM ticket
GROUP BY id;

两个注意事项：
这取决于所用中间管柱长度的内部限制 GROUP_CONAT() （很容易改变的限制）。
它改变了 price 一根绳子。

你可以这样做：

declare @t table (id int, dtsell date, price numeric(18,2),qt int)

insert into @t

values

(1        ,'01-01-2017',   3.00     ,  1),
(1        ,'02-01-2017',   2.00     ,  3),
(2        ,'01-01-2017',   5.00     ,  5),
(2        ,'02-01-2017',   8.00     ,  2)

select x.id,price,z.Qt from (
select id,price,dtsell,row_number() over(partition by id order by dtsell desc ) as rn from @t
)x 
inner join (select SUM(qt) as Qt,ID from @t group by id ) z on x.id = z.id
where rn = 1

您可能需要子查询，请尝试以下操作：

SELECT 
    t1.id,
    (SELECT t2.price FROM ticket t2 WHERE t2.id=t1.id 
       ORDER BY t2.DtSell DESC LIMIT 1 ) AS price, 
    SUM(t1.Qt) 
     FROM ticket t1 GROUP BY t1.id;

您可以从按id分组的ticket中选择所有行（对数量求和），然后连接到每个id组具有最大dtsell的行（选择价格）。
http://sqlfiddle.com/#!9/574断路器9/8

SELECT t.id
         , t3.price
         , SUM(t.Qt)
    FROM   ticket t
    JOIN   ( SELECT t1.id
                  , t1.price
             FROM   ticket t1
             JOIN   ( SELECT id
                           , MAX(dtsell) dtsell
                      FROM   ticket
                      GROUP  BY id ) t2
               ON t1.id      = t2.id
              AND t1.dtsell = t2.dtsell ) t3
      ON t3.id = t.id
    GROUP  BY t.id;