Question

你好，有一个像这样的表：

DateTime               Ip
2016-12-21 17:00:01 |  127.0.0.1 
2016-12-21 17:00:01 |  127.0.0.1 
2016-12-21 17:00:03 |  127.0.0.1 
2016-12-21 17:00:05 |  127.0.0.2 
2016-12-21 17:00:06 |  127.0.0.2 
2016-12-21 17:00:06 |  127.0.0.1 
2016-12-21 17:00:07 |  127.0.0.2 
2016-12-21 17:00:08 |  127.0.0.2 
2016-12-21 17:00:08 |  127.0.0.1 
2016-12-21 17:00:08 |  127.0.0.1

目前在一段时间内计算每秒ip的请求，例如我做的5s：

SELECT Ip, total/diff_in_secs as Rps FROM (
SELECT Ip, count(*) as total, MAX(DateTime), MIN(DateTime), TIMESTAMPDIFF(SECOND, MIN(DateTime), MAX(DateTime)) as diff_in_secs    
FROM requests WHERE DateTime >= '2016-12-21 17:00:01' AND DateTime <= '2016-12-21 17:00:05'
   GROUP BY Ip
   ) as base
   ORDER BY Rps Desc

我试图找出一种方法来在这段时间内做出每个ip的差异和标准差，从rps的任何想法？我试图在这个答案中应用这些概念，但收效甚微Calculate average, variance and standard deviation of two numbers in two different rows/columns with sql / PHP on specific dates

谢谢

Answer 1

试试这个

select 
requests_per_sec.ip, 
avg(requests_per_sec.n) as Avg_Req_Per_Sec,
std(requests_per_sec.n) as Std_Req_Per_Sec
from
 (select ip, count(ip) as n, datetime 
  from requests 
  group by datetime, ip) requests_per_sec 
group by requests_per_sec.ip

要减少内部选择的大小，您还可以使用UNIX_TIMESTAMP(datetime) div 5这样每隔5秒计算一次请求

select 
requests_per_sec.ip, 
avg(requests_per_sec.n) as Avg_Req_Per_Sec,
std(requests_per_sec.n) as Std_Req_Per_Sec
from
 (select ip, count(ip) as n, datetime 
  from requests 
  group by UNIX_TIMESTAMP(datetime) div 5, ip) requests_per_sec 
group by requests_per_sec.ip

如何计算时间范围内的mysql方差和标准差

1 个答案: