选择两行中两个unix时间戳之间的差异

Question

我在MySQL中有一个表格，包含开始和停止时间。

我想锻炼从停止到下一次开始的时间。

我已经做了sqlfiddle我的表 http://sqlfiddle.com/#!2/2c043/1

ID start_time end_time 
410 1367894268 1367898275 
409 1367893164 1367894268 
408 1367888257 1367893153 
407 1367837969 1367888247 

，所以我想要做的是什么

end_time from row 410 - the start_time from row 409 
1367898275 - 1367893164 

和像

end_time  start_time time_duration 
1367898275 1367893164 5111 

感谢您的帮助输出。

Answer 1

TRY像下面，它会帮助你

SQL FIDDLE：http://sqlfiddle.com/#!2/2c043/26

SELECT 
A.idtrip_data, 
A.start_time, 
B.end_time, 
(B.end_time - a.start_time) AS timedifference 
FROM myTable A 
INNER JOIN myTable B 
ON B.idtrip_data = (A.idtrip_data + 1) 
ORDER BY A.idtrip_data ASC 

Answer 2

要实现你问什么尝试以下操作：

SELECT (B.start_time - A.end_time) AS time_duration 
FROM MyTable A INNER JOIN MyTable B ON B.ID = (A.ID + 1) 
ORDER BY A.ID ASC 

IF ID是不连续的，那么你可以使用

SELECT (B.start_time - A.end_time) AS time_duration 
FROM MyTable A CROSS JOIN MyTable B 
WHERE B.ID IN (SELECT MIN (C.ID) FROM MyTable C WHERE C.ID > A.ID) 
ORDER BY A.ID ASC