mysql查询每小时数据和上小时数据的差值实现代码解析

作者:袖梨 2022-06-29

本篇文章小编给大家分享一下mysql查询每小时数据和上小时数据的差值实现代码解析,文章代码介绍的很详细,小编觉得挺不错的,现在分享给大家供大家参考,有需要的小伙伴们可以来看看。

mysql版本:

mysql> select version();
+---------------------+
| version()  |
+---------------------+
| 10.0.22-MariaDB-log |
+---------------------+
1 row in set (0.00 sec)

二、查询每个小时和上小时的差值

1、拆分需求

这里先分开查询下,看看数据都是多少,方便后续的组合。

(1)获取每小时的数据量

这里为了方便展示,直接合并了下,只显示01-12时的数据,并不是bug。。

select count(*) as nums,date_format(log_time,'%Y-%m-%d %h') as days from test where 1 and log_time >='2020-04-19 00:00:00' and log_time <= '2020-04-20 00:00:00' group by days;
+-------+---------------+
| nums | days  |
+-------+---------------+
| 15442 | 2020-04-19 01 |
| 15230 | 2020-04-19 02 |
| 14654 | 2020-04-19 03 |
| 14933 | 2020-04-19 04 |
| 14768 | 2020-04-19 05 |
| 15390 | 2020-04-19 06 |
| 15611 | 2020-04-19 07 |
| 15659 | 2020-04-19 08 |
| 15398 | 2020-04-19 09 |
| 15207 | 2020-04-19 10 |
| 14860 | 2020-04-19 11 |
| 15114 | 2020-04-19 12 |
+-------+---------------+

(2)获取上小时的数据量

select count(*) as nums1,date_format(date_sub(date_format(log_time,'%Y-%m-%d %h'),interval -1 hour),'%Y-%m-%d %h') as days from test where 1 and log_time >='2020-04-19 00:00:00' and log_time <= '2020-04-20 00:00:00' group by days;
+-------+---------------+
| nums1 | days  |
+-------+---------------+
| 15114 | 2020-04-19 01 |
| 15442 | 2020-04-19 02 |
| 15230 | 2020-04-19 03 |
| 14654 | 2020-04-19 04 |
| 14933 | 2020-04-19 05 |
| 14768 | 2020-04-19 06 |
| 15390 | 2020-04-19 07 |
| 15611 | 2020-04-19 08 |
| 15659 | 2020-04-19 09 |
| 15398 | 2020-04-19 10 |
| 15207 | 2020-04-19 11 |
| 14860 | 2020-04-19 12 |
+-------+---------------+

注意:

1)获取上小时数据用的是date_sub()函数,date_sub(日期,interval -1 hour)代表获取日期参数的上个小时,具体参考手册:https://www.w3school.com.cn/sql/func_date_sub.asp

2)这里最外层嵌套了个date_format是为了保持格式和上面的一致,如果不加这个date_format的话,查询出来的日期格式是:2020-04-19 04:00:00的,不方便对比。

2、把这两份数据放到一起看看

select nums ,nums1,days,days1 
from 
(select count(*) as nums,date_format(log_time,'%Y-%m-%d %h') as days from test where 1 and log_time >='2020-04-19 00:00:00' and log_time <= '2020-04-20 00:00:00' group by days) as m,
(select count(*) as nums1,date_format(date_sub(date_format(log_time,'%Y-%m-%d %h'),interval -1 hour),'%Y-%m-%d %h') as days1 from test where 1 and log_time >='2020-04-19 00:00:00' and log_time <= '2020-04-20 00:00:00' group by days1) as n;

+-------+-------+---------------+---------------+
| nums | nums1 | days  | days1  |
+-------+-------+---------------+---------------+
| 15442 | 15114 | 2020-04-19 01 | 2020-04-19 01 |
| 15442 | 15442 | 2020-04-19 01 | 2020-04-19 02 |
| 15442 | 15230 | 2020-04-19 01 | 2020-04-19 03 |
| 15442 | 14654 | 2020-04-19 01 | 2020-04-19 04 |
| 15442 | 14933 | 2020-04-19 01 | 2020-04-19 05 |
| 15442 | 14768 | 2020-04-19 01 | 2020-04-19 06 |
| 15442 | 15390 | 2020-04-19 01 | 2020-04-19 07 |
| 15442 | 15611 | 2020-04-19 01 | 2020-04-19 08 |
| 15442 | 15659 | 2020-04-19 01 | 2020-04-19 09 |
| 15442 | 15398 | 2020-04-19 01 | 2020-04-19 10 |
| 15442 | 15207 | 2020-04-19 01 | 2020-04-19 11 |
| 15442 | 14860 | 2020-04-19 01 | 2020-04-19 12 |
| 15230 | 15114 | 2020-04-19 02 | 2020-04-19 01 |
| 15230 | 15442 | 2020-04-19 02 | 2020-04-19 02 |
| 15230 | 15230 | 2020-04-19 02 | 2020-04-19 03 |

可以看到这样组合到一起是类似于程序中的嵌套循环效果,相当于nums是外层循环,nums1是内存循环。循环的时候先用nums的值,匹配所有nums1的值。类似于php程序中的:

foreach($arr as $k=>$v){
 foreach($arr1 as $k1=>$v1){

 }
}

既然如此,那我们是否可以像平时写程序的那样,找到两个循环数组的相同值,然后进行求差值呢?很明显这里的日期是完全一致的,可以作为对比的条件。

3、使用case …when 计算差值

select (case when days = days1 then (nums - nums1) else 0 end) as diff
from 
(select count(*) as nums,date_format(log_time,'%Y-%m-%d %h') as days from test where 1 and log_time >='2020-04-19 00:00:00' and log_time <= '2020-04-20 00:00:00' group by days) as m,
(select count(*) as nums1,date_format(date_sub(date_format(log_time,'%Y-%m-%d %h'),interval -1 hour),'%Y-%m-%d %h') as days1 from test where 1 and log_time >='2020-04-19 00:00:00' and log_time <= '2020-04-20 00:00:00' group by days1) as n;

效果:
+------+
| diff |
+------+
| 328 |
| 0 |
| 0 |
| 0 |
| 0 |
| 0 |
| 0 |
| 0 |
| 0 |
| 0 |
| 0 |
| 0 |
| 0 |
| -212 |
| 0 |
| 0 

可以看到这里使用case..when实现了当两个日期相等的时候,就计算差值,近似于php程序的:

	foreach($arr as $k=>$v){
 foreach($arr1 as $k1=>$v1){
 if($k == $k1){
  //求差值
 }
 }
}

结果看到有大量的0,也有一部分计算出的结果,不过如果排除掉这些0的话,看起来好像有戏的。

4、过滤掉结果为0 的部分,对比最终数据

这里用having来对查询的结果进行过滤。having子句可以让我们筛选成组后的各组数据,虽然我们的sql在最后面没有进行group by,不过两个子查询里面都有group by了,理论上来讲用having来筛选数据是再合适不过了,试一试

select (case when days = days1 then (nums1 - nums) else 0 end) as diff
from 
(select count(*) as nums,date_format(log_time,'%Y-%m-%d %h') as days from test where 1 and log_time >='2020-04-19 00:00:00' and log_time <= '2020-04-20 00:00:00' group by days) as m,
(select count(*) as nums1,date_format(date_sub(date_format(log_time,'%Y-%m-%d %h'),interval -1 hour),'%Y-%m-%d %h') as days1 from test where 1 and log_time >='2020-04-19 00:00:00' and log_time <= '2020-04-20 00:00:00' group by days1) as n having diff <>0;

结果:
+------+
| diff |
+------+
| -328 |
| 212 |
| 576 |
| -279 |
| 165 |
| -622 |
| -221 |
| -48 |
| 261 |
| 191 |
| 347 |
| -254 |
+------+

这里看到计算出了结果,那大概对比下吧,下面是手动列出来的部分数据:

当前小时和上个小时的差值: 当前小时 -上个小时

本小时上个小时差值

1544215114-328

1523015442212

1465415230576

1493314654-279

1476814933165

可以看到确实是成功获取到了差值。如果要获取差值的比率的话,直接case when days = days1 then (nums1 - nums)/nums1 else 0 end即可。

5、获取本小时和上小时数据的降幅,并展示各个降幅范围的个数

在原来的case..when的基础上引申一下,继续增加条件划分范围,并且最后再按照降幅范围进行group by求和即可。这个sql比较麻烦点,大家有需要的话可以按需修改下,实际测试是可以用的。

select case 
when days = days1 and (nums1 - nums)/nums1 < 0.1 then 0.1
when days = days1 and (nums1 - nums)/nums1 > 0.1 and (nums1 - nums)/nums1 < 0.2 then 0.2
when days = days1 and (nums1 - nums)/nums1 > 0.2 and (nums1 - nums)/nums1 < 0.3 then 0.3
when days = days1 and (nums1 - nums)/nums1 > 0.3 and (nums1 - nums)/nums1 < 0.4 then 0.4
when days = days1 and (nums1 - nums)/nums1 > 0.4 and (nums1 - nums)/nums1 < 0.5 then 0.5
when days = days1 and (nums1 - nums)/nums1 > 0.5 then 0.6
 else 0 end as diff,count(*) as diff_nums
from 
(select count(*) as nums,date_format(log_time,'%Y-%m-%d %h') as days from test where 1 and log_time >='2020-03-20 00:00:00' and log_time <= '2020-04-20 00:00:00' group by days) as m,
(select count(*) as nums1,date_format(date_sub(date_format(log_time,'%Y-%m-%d %h'),interval -1 hour),'%Y-%m-%d %h') as days1 from test where 1 and log_time >='2020-03-20 00:00:00' and log_time <= '2020-04-20 00:00:00' group by days1) as n group by diff having diff >0;

结果:

+------+-----------+

| diff | diff_nums |

+------+-----------+

| 0.1 | 360 |

| 0.2 | 10 |

| 0.3 | 1 |

| 0.4 | 1 |

+------+-----------+

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