一、前言
需求是獲取某個(gè)時(shí)間范圍內(nèi)每小時(shí)數(shù)據(jù)和上小時(shí)數(shù)據(jù)的差值以及比率。本來(lái)以為會(huì)是一個(gè)很簡(jiǎn)單的sql，結(jié)果思考兩分鐘發(fā)現(xiàn)并不簡(jiǎn)單，網(wǎng)上也沒找到參考的方案，那就只能自己慢慢分析了。
剛開始沒思路，就去問DBA同學(xué)，結(jié)果DBA說(shuō)他不會(huì)，讓我寫php腳本去計(jì)算，，這就有點(diǎn)過分了，我只是想臨時(shí)查個(gè)數(shù)據(jù)，就不信直接用sql查不出來(lái)，行叭，咱們邊走邊試。
博主這里用的是笨方法實(shí)現(xiàn)的，各位大佬要是有更簡(jiǎn)單的方式，請(qǐng)不吝賜教，評(píng)論區(qū)等你！
mysql版本：
mysql> select version();
+---------------------+
| version() |
+---------------------+
| 10.0.22-MariaDB-log |
+---------------------+
1 row in set (0.00 sec)
二、查詢每個(gè)小時(shí)和上小時(shí)的差值
1、拆分需求
這里先分開查詢下，看看數(shù)據(jù)都是多少，方便后續(xù)的組合。
（1）獲取每小時(shí)的數(shù)據(jù)量
這里為了方便展示，直接合并了下，只顯示01-12時(shí)的數(shù)據(jù)，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）獲取上小時(shí)的數(shù)據(jù)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）獲取上小時(shí)數(shù)據(jù)用的是date_sub(）函數(shù)，date_sub(日期，interval -1 hour）代表獲取日期參數(shù)的上個(gè)小時(shí)，具體參考手冊(cè)：https://www.w3school.com.cn/sql/func_date_sub.asp
2）這里最外層嵌套了個(gè)date_format是為了保持格式和上面的一致，如果不加這個(gè)date_format的話，查詢出來(lái)的日期格式是：2020-04-19 04:00:00的，不方便對(duì)比。
2、把這兩份數(shù)據(jù)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 |
可以看到這樣組合到一起是類似于程序中的嵌套循環(huán)效果，相當(dāng)于nums是外層循環(huán)，nums1是內(nèi)存循環(huán)。循環(huán)的時(shí)候先用nums的值，匹配所有nums1的值。類似于php程序中的：
foreach($arr as $k=>$v){
foreach($arr1 as $k1=>$v1){
}
}
既然如此，那我們是否可以像平時(shí)寫程序的那樣，找到兩個(gè)循環(huán)數(shù)組的相同值，然后進(jìn)行求差值呢？很明顯這里的日期是完全一致的，可以作為對(duì)比的條件。
3、使用case …when 計(jì)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
可以看到這里使用case..when實(shí)現(xiàn)了當(dāng)兩個(gè)日期相等的時(shí)候，就計(jì)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
}
}
}
結(jié)果看到有大量的0，也有一部分計(jì)算出的結(jié)果，不過如果排除掉這些0的話，看起來(lái)好像有戲的。
4、過濾掉結(jié)果為0 的部分，對(duì)比最終數(shù)據(jù)
這里用having來(lái)對(duì)查詢的結(jié)果進(jìn)行過濾。having子句可以讓我們篩選成組后的各組數(shù)據(jù)，雖然我們的sql在最后面沒有進(jìn)行group by，不過兩個(gè)子查詢里面都有group by了，理論上來(lái)講用having來(lái)篩選數(shù)據(jù)是再合適不過了，試一試
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;
結(jié)果：
+------+
| diff |
+------+
| -328 |
| 212 |
| 576 |
| -279 |
| 165 |
| -622 |
| -221 |
| -48 |
| 261 |
| 191 |
| 347 |
| -254 |
+------+
這里看到計(jì)算出了結(jié)果，那大概對(duì)比下吧，下面是手動(dòng)列出來(lái)的部分?jǐn)?shù)據(jù)：
當(dāng)前小時(shí)和上個(gè)小時(shí)的差值： 當(dāng)前小時(shí) -上個(gè)小時(shí)
本小時(shí) 上個(gè)小時(shí) 差值
15442 15114 -328
15230 15442 212
14654 15230 576
14933 14654 -279
14768 14933 165
可以看到確實(shí)是成功獲取到了差值。如果要獲取差值的比率的話，直接case when days = days1 then (nums1 - nums)/nums1 else 0 end 即可。
5、獲取本小時(shí)和上小時(shí)數(shù)據(jù)的降幅，并展示各個(gè)降幅范圍的個(gè)數(shù)
在原來(lái)的case..when的基礎(chǔ)上引申一下，繼續(xù)增加條件劃分范圍，并且最后再按照降幅范圍進(jìn)行group by求和即可。這個(gè)sql比較麻煩點(diǎn)，大家有需要的話可以按需修改下，實(shí)際測(cè)試是可以用的。
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;
結(jié)果：
+------+-----------+
| diff | diff_nums |
+------+-----------+
| 0.1 | 360 |
| 0.2 | 10 |
| 0.3 | 1 |
| 0.4 | 1 |
+------+-----------+
三、總結(jié)
1、 sql其實(shí)和程序代碼差不多，拆分需求一步步組合，大部分需求都是可以實(shí)現(xiàn)的。一開始就慫了，那自然是寫不出的。
2、 不過復(fù)雜的計(jì)算，一般是不建議用sql來(lái)寫，用程序?qū)憰?huì)更快，sql越復(fù)雜，效率就會(huì)越低。
3、 DBA同學(xué)有時(shí)候也不靠譜，還是要靠自己啊
補(bǔ)充介紹：MySQL數(shù)據(jù)庫(kù)時(shí)間和實(shí)際時(shí)間差8個(gè)小時(shí)
url=jdbc:mysql://127.0.0.1:3306/somedatabase?characterEncoding=utf-8serverTimezone=GMT%2B8
數(shù)據(jù)庫(kù)配置后面加上serverTimezone=GMT%2B8
到此這篇關(guān)于mysql查詢每小時(shí)數(shù)據(jù)和上小時(shí)數(shù)據(jù)的差值的文章就介紹到這了,更多相關(guān)mysql 每小時(shí)數(shù)據(jù)差值內(nèi)容請(qǐng)搜索腳本之家以前的文章或繼續(xù)瀏覽下面的相關(guān)文章希望大家以后多多支持腳本之家！
