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 |
+------+-----------+