计算当前行和前一行之间的时间差

我有我的Web应用程序的下表，我想添加另一列来获取当前行和前一行之间的时间差。我怎样才能实现它？计算当前行和前一行之间的时间差

目前这里是PHP应用程序我的SQL调用

$stmt = $db->prepare("SELECT device,lat,lon, speed, mode, DATE (`currentTime`) ,TIME_FORMAT(`currentTime`, '%H:%i:%s') 
          FROM myTable 
          WHERE device=? limit ?"); 
$stmt ->bind_param('ii', $device_Number ,$limit); 
$stmt ->bind_result($device, $lat, $lon, $speed, $mode, $currentDate, $currentTime); 
$stmt ->execute();

来源

2017-08-14 Achilles

这似乎有点多余存储这样的事情在一个数据库行。你为什么想这样做？ –

我不会将它存储在数据库中，而是在您真正需要时计算它。您可以在数据库端使用https://dev.mysql.com/doc/refman/5.5/zh-CN/date-and-time-functions.html#function_timediff或在PHP中使用DateTime：Diff http：// php .net/manual/de/datetime.diff.php – user1915746

因为我需要知道每行的时差才能进一步计算。 – Achilles

这里，我给样品DATAS有日期时间不同，在这里，你有两种不同的列保存数据所以需要的时间差为2列 '为TimeDifference' 和 'daydifference'

testtime表

id date1  time1 
    1 2017-08-14 01:06:11 
    2 2017-08-14 01:09:13 
    3 2017-08-14 01:16:10 
    4 2017-08-14 01:21:00 
    5 2017-08-15 01:21:00 
    6 2017-08-15 02:13:00

MySQL查询是

SELECT A.id, A.time1, TIMESTAMPDIFF(SECOND,A.time1,B.time1) AS timedifference, 
    TIMESTAMPDIFF(DAY,A.date1,B.date1) AS daydifference 
    FROM testtime A INNER JOIN testtime B ON B.id = (A.id + 1) 
    ORDER BY A.id ASC

来源

2017-08-14 10:05:55 Gopalakrishnan

这假定存在一个id列并且所述列的值是连续的。 – Strawberry

不能让你@Strawberry – Gopalakrishnan

DROP TABLE IF EXISTS my_table; 

CREATE TABLE my_table 
(device INT NOT NULL 
,lat DECIMAL(10,6) NOT NULL 
,lon DECIMAL(10,6) NOT NULL 
,speed DECIMAL(5,2) 
,mode INT NOT NULL 
,dt DATETIME NOT NULL 
,PRIMARY KEY(device,dt) 
); 

INSERT INTO my_table VALUES 
(117,1.415738,103.82360,28.8,3,'2017-07-12 22:07:40'), 
(117,1.424894,103.82561,31.9,3,'2017-07-12 22:08:41'), 
(117,1.429965,103.82674,10.9,3,'2017-07-12 22:09:47'), 
(117,1.430308,103.82873, 5.2,3,'2017-07-12 22:10:47'), 
(117,1.430542,103.83278,13.9,3,'2017-07-12 22:11:48'), 
(117,1.430537,103.83325, 3.2,3,'2017-07-12 22:12:47'); 

SELECT x.* 
    , SEC_TO_TIME(TIME_TO_SEC(x.dt)-TIME_TO_SEC(MAX(y.dt))) diff 
    FROM my_table x 
    LEFT 
    JOIN my_table y 
    ON y.device = x.device 
    AND y.dt < x.dt 
GROUP 
    BY x.device 
    , x.dt; 
+--------+----------+------------+-------+------+---------------------+----------+ 
| device | lat  | lon  | speed | mode | dt     | diff  | 
+--------+----------+------------+-------+------+---------------------+----------+ 
| 117 | 1.415738 | 103.823600 | 28.80 | 3 | 2017-07-12 22:07:40 | NULL  | 
| 117 | 1.424894 | 103.825610 | 31.90 | 3 | 2017-07-12 22:08:41 | 00:01:01 | 
| 117 | 1.429965 | 103.826740 | 10.90 | 3 | 2017-07-12 22:09:47 | 00:01:06 | 
| 117 | 1.430308 | 103.828730 | 5.20 | 3 | 2017-07-12 22:10:47 | 00:01:00 | 
| 117 | 1.430542 | 103.832780 | 13.90 | 3 | 2017-07-12 22:11:48 | 00:01:01 | 
| 117 | 1.430537 | 103.833250 | 3.20 | 3 | 2017-07-12 22:12:47 | 00:00:59 | 
+--------+----------+------------+-------+------+---------------------+----------+ 
6 rows in set (0.00 sec)

来源

2017-08-14 10:07:17 Strawberry

计算当前行和前一行之间的时间差

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