![](/img/trans.png)
[英]MYSQL select field1, field2 only where field1 or field2 does not equal text
[英]mysql correlated subquery : select field1 where max(field2)
选择价格增加最高价的时间最有效的方法是什么? [底部结构]
-获得最高涨价
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)?
-获得最高价格的时间
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
对于数百万个行表,这是一种极其低效的方式,我相信您可以在MSSQL中执行以下操作:
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?
- 结构体 :
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;
我猜这是您想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,您需要一个prices(id, price)
的索引。
最有效的方法是假设id
是顺序的并且没有间隙。 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;
谢谢戈登,当我问这个问题时stackoverflow建议将相关子查询作为标签。 其中撒谎了答案。 因此,这里去:
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。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.