mysql - 本月和上个月两个日期时间之间的平均时间

Question

我们有一个包含两个字段 tkTimeOpen 和 tkTimeClosed 的表。我们需要找出本月和上个月的平均等待时间。我一直无法获得正确的 SQL 查询来提取我需要的内容。

这是记录日期时间的方式； 2017-01-25 10:35

本月平均值；

SELECT SUM(DATEDIFF(MINUTE,tkTimeOpen,tkTimeClose)) * 1.0 
           / (SELECT COUNT(*) * 1.0 FROM e_ticket)
FROM e_ticket
WHERE YEAR(tkTimeOpen) = YEAR(CURRENT_DATE - INTERVAL 1 MONTH) AND MONTH(tkTimeOpen) = MONTH(CURRENT_DATE - INTERVAL 0 MONTH)

上个月的平均值

SELECT SUM(DATEDIFF(MINUTE,tkTimeOpen,tkTimeClose)) * 1.0 
           / (SELECT COUNT(*) * 1.0 FROM e_ticket)
FROM e_ticket
WHERE YEAR(tkTimeOpen) = YEAR(CURRENT_DATE - INTERVAL 1 MONTH) AND MONTH(tkTimeOpen) = MONTH(CURRENT_DATE - INTERVAL 1 MONTH)

我尝试了很多变体，但没有提供所需的输出。 如果有人可以提供帮助，那就太好了!

谢谢

Answer 1

这应该为您提供上个月和本月合并的平均分钟数:

SELECT AVG(TIMESTAMPDIFF(MINUTE, tkTimeOpen, tkTimeClose)) 
FROM e_ticket 
WHERE tkTimeOpen >= DATE_ADD(LAST_DAY(DATE_SUB(NOW(), INTERVAL 2 MONTH)), INTERVAL 1 DAY);

如果您仍然需要分别获得平均值，您可以使您的 WHERE 子句与您拥有的类似...

本月:

SELECT AVG(TIMESTAMPDIFF(MINUTE, tkTimeOpen, tkTimeClose)) 
FROM e_ticket 
WHERE YEAR(tkTimeOpen) = YEAR(NOW()) AND MONTH(tkTimeOpen) = MONTH(NOW());

上个月:

SELECT AVG(TIMESTAMPDIFF(MINUTE, tkTimeOpen, tkTimeClose)) 
FROM e_ticket 
WHERE YEAR(tkTimeOpen) = YEAR(NOW() - INTERVAL 1 MONTH) AND MONTH(tkTimeOpen) = MONTH(NOW() - INTERVAL 1 MONTH);