MySQL 相关子查询:选择 field1，其中 max(field2)

Question

选择价格上涨时间的最有效方法是什么？[底部结构]

-- 获得最大涨价

select p1.pricetime, max(p2.price) maxnext
from prices p1 inner join prices p2 on p2.id > p1.id
group by p1.pricetime

什么是 P2.Pricetime，其中每个 P1.PriceTime 的 P2.Price = Max（P2.price）？

-- 获取最高价格的时间

select p3.pricetime, x.maxnext
from prices p3 inner join 
(select p1.pricetime, max(p2.price) maxnext
from prices p1 inner join prices p2 on p2.id > p1.id
group by p1.pricetime) x
on x.maxnext = p3.price and p3.id > p1.id

对于数百万行表来说，这是一种非常低效的方法。我相信你可以在MSSQL中做这样的事情：

select p2.pricetime from 
(select p1.pricetime, max(p2.price) maxnext
from prices p1 inner join prices p2 on p2.id > p1.id
group by p1.pricetime) x ...

哪个从子查询外部访问子查询别名？

--结构：

CREATE TABLE `prices` (
  `id` int(11) NOT NULL DEFAULT '0',
  `pricetime` varchar(19) DEFAULT NULL,
  `price` decimal(10,8) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
LOCK TABLES `prices` WRITE;
/*!40000 ALTER TABLE `prices` DISABLE KEYS */;
INSERT INTO `prices` (`id`, `pricetime`, `price`)
VALUES
    (1,'2014-01-01 21:55:00',1.37622000),
    (2,'2014-01-01 21:56:00',1.37616000),
    (3,'2014-01-01 21:57:00',1.37616000),
    (4,'2014-01-01 21:58:00',1.37498000),
    (5,'2014-01-01 21:59:00',1.37529000),
    (6,'2014-01-01 22:03:00',1.37518000),
    (7,'2014-01-01 22:05:00',1.37542000),
    (8,'2014-01-01 22:06:00',1.37558000),
    (9,'2014-01-01 22:07:00',1.37560000),
    (10,'2014-01-01 22:08:00',1.37560000);
/*!40000 ALTER TABLE `prices` ENABLE KEYS */;
UNLOCK TABLES;

Answer 1

我猜这是您要获得下一个价格的查询：

select p.*,
       (select p2.price
        from prices p2
        where p2.id > p.id
        order by p2.id
        limit 1
       ) as nextprice
from prices p;

要获得最大的变化，您可以执行以下操作：

select p.*,
       (select p2.price
        from prices p2
        where p2.id > p.id
        order by p2.id
        limit 1
       ) as nextprice
from prices p
order by nextprice - price desc
limit 1;

为了提高性能，您需要在 prices(id, price) 上建立索引。

最有效的方法是假设id是连续的并且没有间隙。 在这种情况下，最好使用自联接：

select p.*, pnext.price
from prices p join
     prices pnext
     on p.id = pnext.id - 1
order by pnext.price - p.price desc
limit 1;

Answer 2

谢谢戈登，当我问问题时，stackoverflow建议将相关子查询作为标签。其中的答案是谎言。所以这里是：

最大增加时间 ：

SELECT p1.pricetime starttime, min(p4.pricetime) endtime, 
p1.price startingprice, p4.price maxnextprice
FROM prices p1 inner join prices p4 on p4.id > p1.id
WHERE p4.price = 
(SELECT max(p3.price) 
FROM prices p2 inner join prices p3 on p3.id > p2.id 
where p1.id = p2.id 
group by p2.pricetime order by max(p3.price) limit 1)
group by p1.pricetime, p4.price;

感谢您的输入。