Question

我们有一个包含两个字段tkTimeOpen和tkTimeClosed的表。我们需要找到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 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)

我尝试了很多变化，但它没有提供所需的输出。如果有人能提供帮助那就太棒了！

谢谢

Answer 1

这应该会给你上个月和本月的平均分钟数：

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

如果你仍然需要单独获得平均值，你可以保持你的WHERE子句与你拥有的相似......

这个月：

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);

Answer 2

您可以使用的一项功能是TIMESTAMPDIFF - 而不是DATEDIFF
您可以使用AVG()代替SUM() / COUNT()

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

本月和上个月的两个日期时间之间的平均时间

2 个答案: