我正在尝试计算和列出网站，以便从一个时间段到下一个时间段的响应时间总体减少最多。
我不需要使用单个查询来执行此操作，我可能会运行多个查询。
网站：
| id | url |
| 1 | stackoverflow.com |
| 2 | serverfault.com |
| 3 | stackexchange.com |
反应：
| id | website_id | response_time | created_at |
| 1 | 1 | 93.26 | 2014-01-28 11:51:39
| 2 | 1 | 99.46 | 2014-01-28 11:52:38
| 2 | 1 | 94.51 | 2014-01-28 11:53:38
| 2 | 1 | 104.46 | 2014-01-28 11:54:38
| 2 | 1 | 85.46 | 2014-01-28 11:56:38
| 2 | 1 | 100.00 | 2014-01-28 11:57:36
| 2 | 1 | 50.00 | 2014-01-28 11:58:37
| 2 | 2 | 100.00 | 2014-01-28 11:58:38
| 2 | 2 | 80 | 2014-01-28 11:58:39
理想情况下，结果如下：
| percentage_change | website_id |
| 52 | 1 |
| 20 | 2 |
我已经找到了最大的响应时间，但不知道如何进行另一个查询来计算最低响应时间，然后进行数学计算，然后对数学进行排序。
SELECT * FROM websites
LEFT JOIN (
SELECT distinct *
FROM responses
ORDER BY response_time desc) responsetable
ON websites.id=responsetable.website_id group by website_id
由于
答案 0 :(得分：5)
您需要等效的lag()或lead()功能。在MySQL中，我使用相关子查询来执行此操作：
select website_id, max(1 - (prev_response_time / response_time)) * 100
from (select t.*,
(select t2.response_time
from table t2
where t2.website_id = t.website_id and
t2.created_at < t.created_at
order by t2.created_at desc
limit 1
) as prev_response_time
from table t
) t
group by website_id;
编辑：
如果您想要从最高到最低的变化：
select website_id, (1 - min(response_time) / max(response_time)) * 100
from table t
group by website_id;
答案 1 :(得分：2)
使用几个序列号： -
SELECT a.id, a.url, MAX(100 * (LeadingResponse.response_time - TrailingResponse.response_time) / LeadingResponse.response_time)
FROM
(
SELECT website_id, created_at, response_time, @aCnt1 := @aCnt1 + 1 AS SeqCnt
FROM responses
CROSS JOIN
(
SELECT @aCnt1:=1
) Deriv1
ORDER BY website_id, created_at
) TrailingResponse
INNER JOIN
(
SELECT website_id, created_at, response_time, @aCnt2 := @aCnt2 + 1 AS SeqCnt
FROM responses
CROSS JOIN
(
SELECT @aCnt2:=2
) Deriv2
ORDER BY website_id, created_at
) LeadingResponse
ON LeadingResponse.SeqCnt = TrailingResponse.SeqCnt
AND LeadingResponse.website_id = TrailingResponse.website_id
INNER JOIN websites a
ON LeadingResponse.website_id = a.id
GROUP BY a.id, a.url
SQL小提琴： -
http://www.sqlfiddle.com/#!2/ace08/1
编辑 - 不同的做法。只有在响应表上的id是按日期/时间顺序时才会有效。
SELECT a.id, a.url, MAX(100 * (r2.response_time - r1.response_time) / r2.response_time)
FROM responses r1
INNER JOIN responses r2
ON r1.website_id = r2.website_id
INNER JOIN
(
SELECT r1.website_id, r1.id, MAX(r2.id) AS prev_id
FROM responses r1
INNER JOIN responses r2
ON r1.website_id = r2.website_id
AND r1.id > r2.id
GROUP BY r1.website_id, r1.id
) ordering_query
ON r1.website_id = ordering_query.website_id
AND r1.id = ordering_query.id
AND r2.id = ordering_query.prev_id
INNER JOIN websites a
ON r1.website_id = a.id
GROUP BY a.id, a.url
您可以根据response_time字段而不是id执行类似的操作，但这需要网站的response_time是唯一的。
修改
刚刚看到你不仅想要连续的变化，而只需要从最高到最低的响应。假设最低点不必在最高点之后出现： -
SELECT id, url, MAX(100 * (max_response - min_response) / max_response)
FROM
(
SELECT a.id, a.url, MIN(r1.response_time) AS min_response, MAX(r1.response_time) AS max_response
FROM responses r1
INNER JOIN websites a
ON r1.website_id = a.id
GROUP BY a.id, a.url
) Sub1
如果您只对较高响应时间之后的较低响应时间感兴趣： -
SELECT id, url, MAX(100 * (max_response - min_following_response) / max_response)
FROM
(
SELECT a.id, a.url, MAX(r1.response_time) AS max_response, MIN(r2.response_time) AS min_following_response
FROM responses r1
INNER JOIN responses r2
ON r1.website_id = r2.website_id
AND (r1.created_at < r2.created_at
OR (r1.created_at = r2.created_at
AND r1.id < r2.id))
INNER JOIN websites a
ON r1.website_id = a.id
GROUP BY a.id, a.url
) Sub1
（假设响应表上的id字段是唯一的并且按顺序创建）
答案 2 :(得分：1)
从你的“我已经找到最大的响应时间，但不知道如何进行另一个查询来计算最低的响应时间，然后进行数学运算，然后对数学进行排序。”我明白你想要最短的响应时间和最大的响应时间并做数学。
drop table #test
create table #test(
id int, website_id int, response_time decimal, created_at datetime)
insert into #test (id , website_id , response_time , created_at) values ( 1 , 1 , 93.26, '2014-01-28 11:51:39')
insert into #test (id , website_id , response_time , created_at) values ( 2 , 1 , 99.46 , '2014-01-28 11:52:38')
insert into #test (id , website_id , response_time , created_at) values ( 2 , 1 , 94.51 , '2014-01-28 11:53:38')
insert into #test (id , website_id , response_time , created_at) values ( 2 , 1 , 104.46 , '2014-01-28 11:54:38')
insert into #test (id , website_id , response_time , created_at) values ( 2 , 1 , 85.46 , '2014-01-28 11:56:38')
insert into #test (id , website_id , response_time , created_at) values ( 2 , 1 , 100.00 , '2014-01-28 11:57:38')
insert into #test (id , website_id , response_time , created_at) values ( 2 , 1 , 50.00 , '2014-01-28 11:58:38')
insert into #test (id , website_id , response_time , created_at) values ( 2 , 2 , 100.00 , '2014-01-28 11:58:38')
insert into #test (id , website_id , response_time , created_at) values ( 2 , 2 , 80 , '2014-01-28 11:58:38')
select * from #test
select distinct MINT.website_id,MINT.MINRT,maxT.MINRT,(MINT.MINRT/maxT.MINRT)*100--Do your calculation here---
from #test t0
inner join(select min(response_time) as MINRT,website_id from #test group by website_id ) MINT
on MINT.website_id = t0.website_id
inner join(select max(response_time) as MINRT,website_id from #test group by website_id ) maxT
on maxT.website_id = t0.website_id
答案 3 :(得分：1)
您想将最短响应时间除以每个网站的最长响应时间吗？那只是：
select
websites.id as website_id,
100 - min(response_time) / max(response_time) * 100 as percentage_change
from websites
left join responses on websites.id = responses.website_id
group by websites.id;
（我假设response_time永远不会为零。如果可以的话，你将不得不使用case语句。）
答案 4 :(得分：0)
按website_id对响应时间进行分组，找到MIN(response_time)和MAX(response_time)，并比较MIN()后MAX()是否仅过滤了可提高其效果的网站。
<?php
$rows = $db->fetchAll('
select
r.website_id, min(r.response_time) min_time, max(r.response_time) max_time,
(select
rmin.created_at
FROM
responses rmin
WHERE
rmin.response_time = min(r.response_time) AND
rmin.website_id = r.website_id
ORDER BY rmin.created_at
LIMIT 1) min_created_at,
(select
rmax.created_at
FROM
responses rmax
WHERE
rmax.response_time = max(r.response_time) AND
rmax.website_id = r.website_id
ORDER BY rmax.created_at DESC
LIMIT 1) max_created_at
FROM
responses r
GROUP BY
r.website_id');
foreach($rows as $row) {
if($row['max_created_at'] < $row['min_created_at']) {
echo 'Website '.$row['website_id'].
' improved by '.
(100 - (($row['min_time'] / $row['max_time']) * 100)).
"%\n";
}
}
然而，对于大型数据集，查询可能很慢。您必须优化索引和/或查询 sqlfiddle：http://www.sqlfiddle.com/#!2/fa8f9/8