MySQL减法和求和

问题描述

我在mysql中有一张下表。

DROP TABLE IF EXISTS `mdc_data`;

CREATE TABLE `mdc_data` (
`id` int(11) NOT NULL AUTO_INCREMENT,`tariff_id` int(11) DEFAULT NULL,`msn` varchar(100) DEFAULT NULL,`cust_id` varchar(100) DEFAULT NULL,`kwh_t_max` varchar(100) DEFAULT NULL,`start_data_date_time` datetime DEFAULT NULL,`end_data_date_time` datetime DEFAULT NULL,`incoming_source` varchar(100) DEFAULT NULL,`t_type` varchar(100) DEFAULT NULL,PRIMARY KEY (`id`)
 ) ENGINE=InnoDB AUTO_INCREMENT=15 DEFAULT CHARSET=latin1;

 /*Data for the table `mdc_data` */

insert  into `mdc_data`(`id`,`tariff_id`,`msn`,`cust_id`,`kwh_t_max`,`start_data_date_time`,`end_data_date_time`,`incoming_source`,`t_type`) values (1,12,'00209706','37010114710','281','2020-09-10 00:40:00','2020-09-10 01:10:00','WAPDA','Slab Based'),(2,'00209707','37010114712','103',(3,'297','2020-09-10 02:00:00','2020-09-10 02:30:00',(4,'119',(5,'312','2020-09-10 03:20:00','2020-09-10 03:50:00',(6,'135',(7,'285','2020-09-10 01:20:00','2020-09-10 01:50:00','Generator',(8,'107',(9,'304','2020-09-10 02:40:00','2020-09-10 03:10:00',(10,'127',(11,'320','2020-09-10 04:00:00','2020-09-10 04:30:00',(12,'143','2020-09-10 04:40:00','2020-09-10 05:10:00',(13,'268','2020-09-09 23:20:00','2020-09-10 00:00:00',(14,'95','Slab Based');

我的SQL Fiddle

通过使用此查询SELECT * FROM mdc_data d WHERE d. msn ='00209706' ORDER BY d. start_data_date_time ASC我得到了

 id|tariff_id|msn     | cust_id   |kwh_t_max|start_data_date_time|end_data_date_time |incoming_source|t_type    |
=================================================================================================================
 13|    12   |00209706|37010114710|    268  |2020-09-09 23:20:00 |2020-09-10 00:00:00|Generator      |Slab Based|
  1|    12   |00209706|37010114710|    281  |2020-09-10 00:40:00 |2020-09-10 01:10:00|WAPDA          |Slab Based|
  7|    12   |00209706|37010114710|    285  |2020-09-10 01:20:00 |2020-09-10 01:50:00|Generator      |Slab Based|
  3|    12   |00209706|37010114710|    297  |2020-09-10 02:00:00 |2020-09-10 02:30:00|WAPDA          |Slab Based|
  9|    12   |00209706|37010114710|    304  |2020-09-10 02:40:00 |2020-09-10 03:10:00|Generator      |Slab Based|
  5|    12   |00209706|37010114710|    312  |2020-09-10 03:20:00 |2020-09-10 03:50:00|WAPDA          |Slab Based|
 11|    12   |00209706|37010114710|    320  |2020-09-10 04:00:00 |2020-09-10 04:30:00|Generator      |Slab Based|
....

我想做什么？

2020-09-10 00:00:00处的第一个值是起始值。

我想按每米计出消耗量。例如，仪表号是00209706。为了获得WAPDA的消费量，需要以下算法。

281 - 268 = 13 | 297 - 285 = 12 | 312 - 304 = 8  |
// Adding the difference value 
13+12+8 = 33 => is the consumption against `WAPDA`

类似于发电机

 285 - 281 = 4 | 304 - 297 = 7 | 320 - 312 = 8  |
// Adding the difference value 
4+7+8 = 19=> is the consumption against `Generator`

预期产量

 msn    | units | incoming_source | date       |    
====================================================
00209706| 33    | WAPDA           | 2020-09-10 |
00209706| 19    | Generator       | 2020-09-10 | 

....

解决方法

我认为您想要lag()，然后进行汇总：

select msn,sum(kwh_t_max - lag_kwh_t_max) units,incoming_source,date(end_data_date_time) date
from (
    select 
        t.*,lag(kwh_t_max) over(partition by msn order by end_data_date_time) lag_kwh_t_max
    from mytable t
) t
group by msn,date(end_data_date_time)

我发现将消耗量减去属于相同来源的行会更有意义-但这就是您要的。

Demo on DB Fiddle ：

msn      | units | incoming_source | date      
:------- | ----: | :-------------- | :---------
00209706 |    19 | Generator       | 2020-09-10
00209706 |    33 | WAPDA           | 2020-09-10
00209707 |    20 | Generator       | 2020-09-10
00209707 |    28 | WAPDA           | 2020-09-10