我有一个问题。

select contract_no AS c_no, cm_mac AS c_mc, MIN(tstamp) as time2, sum(1) as aps
from devices where 
contract_no in 
(select distinct(contract_no) from devices where 
tstamp >= '2018-10-28 06:59:59' AND tstamp <= '2018-10-29 07:00:00')
group by contract_no, cm_mac;

我意识到查询速度很慢，所以我想知道是否有可能优化这个查询？我在想也许用exists代替in，但在这种情况下我不能从exists开始 EXISTS (SELECT 1 from .... where contract_no= contract_no ) 因为我需要这个 DISTINCT 条款。
当然，我需要返回相同的结果。这是否可能以某种方式优化此查询？

UPDATE:

我查看了反馈，你是对的。如果执行这两个查询，我将得到相同的结果。但关键是完整的查询更复杂，如果没有这个子查询，我会得到更多的结果。
查询1（返回正确的72行）：

SELECT id, contract_no, customer, address, cm_mac, aps 
    FROM (select * from new_installed_devices where  insert4date >='2018-10-28' 
    AND insert4date <='2018-10-28' AND install_mark<2) as d1 
left join 
( select * from (select contract_no AS c_no, cm_mac AS c_mc, 
MIN(tstamp) as time2, sum(1) as aps from devices_change 
where contract_no in (select distinct(contract_no) from devices_change 
where tstamp >= '2018-10-28 06:59:59' AND tstamp <= '2018-10-29 07:00:00') 
group by contract_no, cm_mac ) as mtmbl 
where mtmbl.time2 >= '2018-10-28 06:59:59' and mtmbl.time2 <= '2018-10-29 
07:00:00' ) as tmp on d1.contract_no=tmp.c_no 
where aps>0 group by contract_no, customer, address, cm_mac;

查询2（返回75行不正确），您建议使用此方法（在一行中包含两个查询）：

SELECT id, contract_no, customer, address, cm_mac, aps  
FROM (select * from new_installed_devices where  insert4date >='2018-10-28' 
AND insert4date <='2018-10-28' AND install_mark<2) as d1 left join 
( select * from (select distinct(contract_no) AS c_no, cm_mac AS c_mc, 
MIN(tstamp) as time2, sum(1) as aps from devices_change 
where  tstamp >= '2018-10-28 06:59:59' AND tstamp <= '2018-10-29 07:00:00'
 group by contract_no, cm_mac ) as mtmbl 
where mtmbl.time2 >= '2018-10-28 06:59:59' and 
mtmbl.time2 <= '2018-10-29 07:00:00' ) as tmp 
on d1.contract_no=tmp.c_no 
where aps>0 group by contract_no, customer, address, cm_mac;