我在flights之间旅行了很多airports。
每个机场都有一个ID和（x，y）坐标。
对于属于用户的特定航班列表，我想找到所访问的最北端（最高y）机场。
以下是我目前正在使用的查询：
SELECT name,iata,icao,apid,x,y
FROM airports
WHERE y=(SELECT MAX(y)
FROM airports AS a
, flights AS f
WHERE (f.src_apid=a.apid OR f.dst_apid=a.apid) AND f.uid=[user_id]
)
只要y是唯一的（=在那个纬度上只有一个机场），这种方法运行得非常好而且速度相当快，但是一旦它没有，就会失败。不幸的是，这经常发生，例如。军用和民用机场即使占据相同的坐标也有单独的条目。
我真正想做的是在子查询中找到MAX(y)的机场并返回实际匹配的机场（a.apid），而不是返回y的值和然后再次匹配。有什么建议吗？
假设用户只有这一个航班，来自apid'3728'：
mysql> select * from flights where uid=35 and src_apid=3728 limit 1;
+------+----------+----------+----------+----------+------+------+-----------+-------+--------+------+------+------+--------+----------+--------------+--------------+---------------------+------+------------+------+
| uid | src_apid | src_time | dst_apid | distance | code | seat | seat_type | class | reason | plid | alid | trid | fid | duration | registration | note | upd_time | opp | src_date | mode |
+------+----------+----------+----------+----------+------+------+-----------+-------+--------+------+------+------+--------+----------+--------------+--------------+---------------------+------+------------+------+
| 35 | 3728 | NULL | 3992 | 4116 | NW16 | 23C | A | Y | L | 167 | 3731 | NULL | 107493 | 08:00:00 | | del. typhoon | 2008-10-04 10:40:58 | Y | 2001-08-22 | F |
+------+----------+----------+----------+----------+------+------+-----------+-------+--------+------+------+------+--------+----------+--------------+--------------+---------------------+------+------------+------+
在同一坐标处有两个机场：
mysql> select * from airports where y=21.318681;
+-----------------------+----------+---------------+------+------+-------------+-----------+-----------+------+------+----------+------+
| name | city | country | iata | icao | x | y | elevation | apid | uid | timezone | dst |
+-----------------------+----------+---------------+------+------+-------------+-----------+-----------+------+------+----------+------+
| Honolulu Intl | Honolulu | United States | HNL | PHNL | -157.922428 | 21.318681 | 13 | 3728 | NULL | -10 | N |
| Hickam Air Force Base | Honolulu | United States | | PHIK | -157.922428 | 21.318681 | 13 | 7055 | 3 | -10 | N |
+-----------------------+----------+---------------+------+------+-------------+-----------+-----------+------+------+----------+------+
如果你运行原始查询，子查询将返回y = 21.318681，而后者将匹配apid 3728（正确）或 apid 7055（错误）。
答案 0 :(得分：2)
这个怎么样：
SELECT name,iata,icao,apid,x,y
FROM airports AS a, flights AS f
WHERE f.src_apid=a.apid OR f.dst_apid=a.apid
ORDER BY y DESC LIMIT 1
您将从相关用户的所有航班起飞，从北到南订购，并从列表中选择第一个。
答案 1 :(得分：1)
第三次尝试，使用假定的用户（userid，name）表
select u.name, ap.name
, ap.iata
, ap.icao
, ap.apid
, ap.x
, max(ap.y)
from users u
, airports ap
, flights f
where u.userid=f.userid
and ( f.src_apid=ap.apid
OR f.dst_apid=ap.apid
)
group by u.name, ap.name,ap.iata,ap.icao,ap.apid,ap.x
现在，您可以将查询限制为您感兴趣的用户..
评论GROUP BY：
答案 2 :(得分：1)
以下查询如何执行？ 它的工作原理是首先找到所访问的机场中最北端的Y坐标。然后执行相同的查询，该查询由先前查询中的Y坐标过滤。最后一步是找到机场。
drop table airports;
drop table flights;
create table airports(
apid int not null
,apname varchar(50) not null
,x int not null
,y int not null
,primary key(apid)
,unique(apname)
);
create table flights(
flight_id int not null auto_increment
,src_apid int not null
,dst_apid int not null
,user_id varchar(20) not null
,foreign key(src_apid) references airports(apid)
,foreign key(dst_apid) references airports(apid)
,primary key(flight_id)
,index(user_id)
);
insert into airports(apid, apname, x, y) values(1, 'Northpole Civilian', 50, 100);
insert into airports(apid, apname, x, y) values(2, 'Northpole Military', 50, 100);
insert into airports(apid, apname, x, y) values(3, 'Transit point', 50, 50);
insert into airports(apid, apname, x, y) values(4, 'Southpole Civilian', 50, 0);
insert into airports(apid, apname, x, y) values(5, 'Southpole Military', 50, 0);
insert into flights(src_apid, dst_apid, user_id) values(4, 3, 'Family guy');
insert into flights(src_apid, dst_apid, user_id) values(3, 1, 'Family guy');
insert into flights(src_apid, dst_apid, user_id) values(5, 3, 'Mr Bazooka');
insert into flights(src_apid, dst_apid, user_id) values(3, 2, 'Mr Bazooka');
select airports.apid
,airports.apname
,airports.x
,airports.y
from (select max(a.y) as y
from flights f
join airports a on (a.apid = f.src_apid or a.apid = f.dst_apid)
where f.user_id = 'Family guy'
) as northmost
join (select a.apid
,a.y
from flights f
join airports a on (a.apid = f.src_apid or a.apid = f.dst_apid)
where f.user_id = 'Family guy'
) as userflights on(northmost.y = userflights.y)
join airports on(userflights.apid = airports.apid);
修改。替代查询可能不会让优化程序混淆
select airports.*
from (select case when s.y > d.y then s.apid else d.apid end as apid
,case when s.y > d.y then s.y else d.y end as northmost
from flights f
join airports s on(f.src_apid = s.apid)
join airports d on(f.dst_apid = d.apid)
where f.user_id = 'Family guy'
order by northmost desc
limit 1
) as user_flights
join airports on(airports.apid = user_flights.apid);
答案 3 :(得分：1)
好的，也许是这样的：
SELECT name, iata, icao, apid, x, y
FROM airports
WHERE y = (SELECT MAX(A.y)
FROM airports AS a
INNER JOIN flights AS f
ON (F.SRC_APID = A.APID OR
F.DST_APID = A.APID)
WHERE f.uid = [user_id]) AND
apid IN (SELECT SRC_APID AS APID
FROM FLIGHTS
WHERE UID = [user_id]
UNION ALL
SELECT DEST_APID AS APID
FROM FLIGHTS
WHERE UID = [user_id])
无法保证这将如何表现，但也许这是朝着正确方向迈出的一步。
分享并享受。