我将事件记录存储在一个表中,开始和结束日期时间记录为单独的记录。
存储记录的表。
CREATE TABLE `avl_data` (
`id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
`imei_number` bigint(20) unsigned NOT NULL DEFAULT '0',
`latitude` double NOT NULL DEFAULT '0',
`longitude` double NOT NULL DEFAULT '0',
`report_id` tinyint(4) NOT NULL DEFAULT '0',
`rtc_date` datetime NOT NULL,
`ibutton_id` varchar(25) DEFAULT 'N/A',
`odometer` bigint(20) NOT NULL DEFAULT '0',
`speed` smallint(6) NOT NULL DEFAULT '0',
`vin_number` varchar(255) DEFAULT 'N/A',
`date_created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`time_report` int(11) NOT NULL DEFAULT '0',
PRIMARY KEY (`id`),
KEY `imei_number` (`imei_number`),
KEY `imei_rtc` (`imei_number`,`rtc_date`),
CONSTRAINT `avl_data_ibfk_1` FOREIGN KEY (`imei_number`) REFERENCES `assets` (`imei_number`) ON DELETE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=64916466 DEFAULT CHARSET=utf8 |`
这就是我迄今为止所尝试的。
select concat(ass.label_1, " ", ass.label_2, " ", ass.label_3) as "Vehicle",
@start := case
when a.report_id = 103 then convert_tz(a.rtc_date, 'UTC', 'Asia/Kuwait') end as "Start",
@end := case
when a.report_id = 104 then convert_tz(a.rtc_date, 'UTC', 'Asia/Kuwait') end as "End",
TIMEDIFF(@start, @end) as 'Duration',
a.speed as 'Speed',
a.latitude as 'Latitude',
a.longitude as 'Longitude'
from avl_data a
inner join assets ass on a.imei_number = ass.imei_number
where a.imei_number = 356158069811103
and rtc_date >= '2018-10-01 00:00:00'
and rtc_date <= '2018-10-31 23:59:59'
and a.report_id in (103, 104)
order by a.rtc_date asc;
它分别产生开始和事件记录,第一个记录是事件开始,第二个记录是事件停止。
+-------------------------------+---------------------+---------------------+----------+-------+-----------+-----------+
| Vehicle | Start | End | Duration | Speed | Latitude | Longitude |
+-------------------------------+---------------------+---------------------+----------+-------+-----------+-----------+
| Mitsubishi Outlander 14/74080 | 2018-10-01 08:29:26 | NULL | NULL | 128 | 29.045856 | 48.113764 |
| Mitsubishi Outlander 14/74080 | NULL | 2018-10-01 08:30:17 | NULL | 114 | 29.031169 | 48.121516 |
|
理想情况下,我想要一排,即
+-------------------------------+---------------------+---------------------+----------+-------+-----------+-----------+
| Vehicle | Start | End | Duration | Speed | Latitude | Longitude |
+-------------------------------+---------------------+---------------------+----------+-------+-----------+-----------+
| Mitsubishi Outlander 14/74080 | 2018-10-01 08:29:26 | 2018-10-01 08:30:17 | 00:01:17 | 128 | 29.045856 | 48.113764 |
多亏了@thorsten,这项工作对我起了作用,领导职能发挥得非常好。
select a.report_id as "ID",
any_value(case when a.report_id = 103 then concat(ass.label_1, ' ', ass.label_2, ' ', ass.label_3) end) as "Vehicle",
any_value(case when a.report_id = 103 then convert_tz(a.rtc_date, 'UTC', 'Asia/Kuwait') end) as "Start",
any_value(case when a.report_id = 103 then lead(convert_tz(a.rtc_date, 'UTC', 'Asia/Kuwait')) over () end) as "End",
any_value(case when a.report_id = 103 then SEC_TO_TIME(TIMESTAMPDIFF(SECOND , convert_tz(a.rtc_date, 'UTC', 'Asia/Kuwait'), lead(convert_tz(a.rtc_date, 'UTC', 'Asia/Kuwait')) over ())) end) as "Duration",
any_value(case when a.report_id = 103 then a.speed end) as "Speed",
any_value(case when a.report_id = 103 then a.latitude end) as "Latitude",
any_value(case when a.report_id = 103 then a.longitude end) as "Longitude"
from avl_data a
join assets ass on a.imei_number = ass.imei_number
where a.imei_number = 356158069811103
and a.rtc_date >= '2018-10-01 00:00:00'
and a.rtc_date <= '2018-10-31 23:59:59'
and a.report_id in (103, 104);
结果,尽管有办法删除现在的空行吗?
+-----+-------------------------------+---------------------+---------------------+----------+-------+-----------+-----------+
| ID | Vehicle | Start | End | Duration | Speed | Latitude | Longitude |
+-----+-------------------------------+---------------------+---------------------+----------+-------+-----------+-----------+
| 103 | Mitsubishi Outlander 14/74080 | 2018-10-01 08:29:26 | 2018-10-01 08:30:17 | 00:00:51 | 128 | 29.045856 | 48.113764 |
| 104 | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
| 103 | Mitsubishi Outlander 14/74080 | 2018-10-01 08:43:45 | 2018-10-01 08:44:14 | 00:00:29 | 136 | 29.067756 | 48.110384 |
| 104 | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
| 103 | Mitsubishi Outlander 14/74080 | 2018-10-01 09:31:36 | 2018-10-01 09:31:44 | 00:00:08 | 135 | 29.056563 | 48.108851 |
| 104 | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
| 103 | Mitsubishi Outlander 14/74080 | 2018-10-01 09:32:02 | 2018-10-01 09:33:54 | 00:01:52 | 149 | 29.048803 | 48.112581 |
| 104 | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
| 103 | Mitsubishi Outlander 14/74080 | 2018-10-01 09:41:57 | 2018-10-01 09:42:35 | 00:00:38 | 131 | 29.036886 | 48.108733 |
| 104 | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
| 103 | Mitsubishi Outlander 14/74080 | 2018-10-01 19:48:09 | 2018-10-01 19:48:20 | 00:00:11 | 126 | 29.034386 | 48.119706 |
| 104 | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
2条答案
按热度按时间o75abkj41#
这两行之间有公用键吗?
如果是这样的话,那么只需将avl\ U数据表的两个示例连接在一起
像这样的
其中key是公共密钥。
如果您没有一个公共键,并且依赖一行接一行,那么您有一个主键序列,那么类似的方法可能会起作用
但是,我不建议这样做,因为对于整个表来说,一个id比前一个id大一个可能不是100%可靠的。
像这样的东西可能会给你答案,但它会有额外的结果,因为它会有一个开始与结束以及
您可以将其用作子查询,并用排名值ref Package 它https://stackoverflow.com/a/1895127/3805124 然后删除偶数,因为它们应该是以结束记录开始的多余的。
vlf7wbxs2#
可以使用聚合从两行中获取一行。当您需要一些来自report\u id=103的值和一些来自report\u id=104的值时,您可以使用
CASE WHEN
得到一个或另一个值。由于只有一个103行和一个104行,所以聚合函数是一个伪聚合,具有ANY_VALUE
.另一种选择是加入: