


How to calculate the difference between two adjacent rows in a certain column in MySQL
MySQL method of calculating the difference between two adjacent rows in a column: first use [r1.rownum = r2.rownum - 1] to determine whether the two records are the previous and next rows; then use the TIMEDIFF function to calculate the time difference That’s it.
【Related learning recommendations: mysql tutorial(Video)】
MySQL method for calculating the difference in a column between two adjacent rows:
First of all, the blogger has a table on the server to record the GPS reported by the driver. Point information, the table structure is as follows:
-- 司机GPS收集表 CREATE TABLE captainad_driver_gps_position ( id BIGINT NOT NULL auto_increment COMMENT '主键', business_id BIGINT DEFAULT NULL COMMENT '业务ID', device_mac VARCHAR (64) DEFAULT NULL COMMENT '设备MAC地址', device_imei VARCHAR (64) DEFAULT NULL COMMENT '设备IMEI', lat_lng VARCHAR (64) DEFAULT NULL COMMENT '纬经度', capture_time TIMESTAMP DEFAULT CURRENT_TIMESTAMP COMMENT '捕获时间', create_time TIMESTAMP DEFAULT CURRENT_TIMESTAMP COMMENT '创建时间', update_time TIMESTAMP DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP COMMENT '修改时间', PRIMARY KEY (id), KEY `idx_business_id` (`business_id`) USING BTREE ) ENGINE = INNODB DEFAULT CHARSET = utf8 COMMENT = '司机GPS收集';
The data recorded in the table is roughly as follows:
Now, after sorting by time, capture_time, the time to obtain the GPS position, make two records before and after Calculate the difference. In order to calculate the difference between the two, we definitely need to obtain two records, one before the other and one after the other. Here we can cleverly use a variable to record the number of rows in the current row, and then superimpose the number of rows each time with the loop query. , to achieve the purpose of row recording, so that we can know which two records are one after another.
SQL statement to print row number:
SELECT (@rownum := @rownum + 1) AS rownum, tab.business_id, tab.device_mac, tab.capture_time FROM captainad_driver_gps_position tab, (SELECT @rownum := 0) r -- 声明变量 WHERE 1 = 1 AND DATE_FORMAT( tab.capture_time, '%Y-%m-%d' ) = '2019-06-28' ORDER BY tab.capture_time
Based on this, we write the target SQL. Here I have slightly organized the statement based on our actual business. The script is roughly as follows:
SELECT t.business_id, t.device_mac, t.capture_time, t.tdiff FROM ( SELECT r1.business_id, r1.device_mac, r1.capture_time, TIMEDIFF( r2.capture_time, r1.capture_time ) AS 'tdiff' FROM ( SELECT (@rownum := @rownum + 1) AS rownum, tab.business_id, tab.device_mac, tab.capture_time FROM captainad_driver_gps_position tab, (SELECT @rownum := 0) r WHERE 1 = 1 AND DATE_FORMAT( tab.capture_time, '%Y-%m-%d' ) = '2019-06-28' ORDER BY tab.capture_time ) r1 LEFT JOIN ( SELECT (@INDEX := @INDEX + 1) AS rownum, tab.business_id, tab.device_mac, tab.capture_time FROM captainad_driver_gps_position tab, (SELECT @INDEX := 0) r WHERE 1 = 1 AND DATE_FORMAT( tab.capture_time, '%Y-%m-%d' ) = '2019-06-28' ORDER BY tab.capture_time ) r2 ON r1.business_id = r2.business_id AND r1.device_mac = r2.device_mac AND r1.rownum = r2.rownum - 1 ) t WHERE t.tdiff > '00:00:15'
In the above code, we use r1.rownum = r2.rownum - 1
to determine whether the two records are the previous and next rows, and then use the TIMEDIFF
function Let’s calculate the time difference. At this point, our goal has been achieved.
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