查询中产生的多次求和 (timediff)

Question

我有两个字段：

初始（时间戳）
最终（时间戳）

我的查询是：

SELECT TIMEDIFF(Final, Initial) 
    AS 'Worked Hours' 
    FROM `db_foo`
    WHERE matriculation='X' AND date='2017-yy-yy'

结果会是这样的

Worked Hours    
03:34:00
02:34:00
01:00:00
[...]

是否可以将这些多个时间戳进一步相加，从而得到总工作时间？

示例数据集（以 csv 格式导出）：

DATE --- ID --- INITIAL --- FINAL --- MATRICULATION

2017-09-14,"29","2017-09-14 11:00:00","2017-09-14 14:34:00","4"
2017-09-14,"30","2017-09-14 17:00:00","2017-09-14 19:34:00","4"
2017-09-14,"31","2017-09-14 21:00:00","2017-09-14 22:00:00","4"

期望的输出（它是工作时间的总和）：

Worked Hours
07:08:00

提前致谢

Answer 1

要获得所需的结果，您可以使用以下查询

SELECT SEC_TO_TIME(
  SUM(
    TIMESTAMPDIFF(SECOND,Initial,Final)
  )
)
FROM `db_foo` /* WHERE clause*/;

要获得总和以及之前的结果集，您可以按照以下方法

SELECT t.*,SEC_TO_TIME(SUM(workedhours))
FROM (
  SELECT ID, TIMESTAMPDIFF(SECOND,Initial,Final)  workedhours
  FROM `db_foo`  /* WHERE clause*/
 ) t
GROUP BY ID WITH ROLLUP;

DEMO

Answer 2

TIME 类型的最大值为 838:59:59。如果您认为总和可能超过 838 小时，则对 TIME 表达式求和是不安全的。我建议将时差转换为分钟并将总小时数显示为十进制数，而不是时间：

SELECT
  ROUND(SUM(TIMESTAMPDIFF(MINUTE, Initial, Final) / 60.0), 1) AS "Worked Hours"
FROM `db_foo`
WHERE matriculation='X' AND date='2017-yy-yy';

这会 return

Worked Hours
7.1

Answer 3

试试这个

用过SEC_TO_TIME and TIME_TO_SEC。

SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(TIMEDIFF(Final, Initial))))
AS 'Worked Hours' 
FROM `db_foo`
WHERE matriculation='X' AND date='2017-yy-yy'

希望对你有所帮助

查询中产生的多次求和 (timediff)

Sum multiple times produced in a query (timediff)

mysql

sql-timestamp