如何在几周内对mysql结果进行分组

Question

我有一张这样的表：

我需要总结从输入日期开始的过去8周（但每周）每个msisdn发送的消息数量。以下是我提出的建议：

SELECT count(*) as ukupan_broj, SUM(IF (sent_messages.delivered = 1,1,0 )) as broj_dostavljenih, 
count(*) - SUM(IF (sent_messages.delivered = 1,1,0 )) as non_billed,
SUM(IF (sent_messages.delivered = 1,1,0 )) / count(*) as ratio,
`sent_messages`.`msisdn`, 
MONTH(`sent_messages`.`datetime`) AS MONTH, WEEK(`sent_messages`.`datetime`) AS WEEK,
DATE_FORMAT(`sent_messages`.`datetime`, '%Y-%m-%d') AS DATE
FROM `sent_messages`
INNER JOIN `received_messages` on `received_messages`.`uniqueid`=`sent_messages`.`originalID` 
and `received_messages`.`msisdn`=`sent_messages`.`msisdn`
WHERE `sent_messages`.`datetime` >= '2016-12-12'
AND `sent_messages`.`originalID` = `received_messages`.`uniqueid`
AND `sent_messages`.`datetime` <= '2017-12-30'
AND  `sent_messages`.`datetime` >=  `received_messages`.`datetime`
AND  `sent_messages`.`datetime` <= ( `received_messages`.`datetime` + INTERVAL 2 HOUR )
AND `sent_messages`.`type` = 'PAID'
GROUP BY WEEK
ORDER BY  DATE ASC

因为我按WEEK对其进行分组，我的结果是显示所有已发送，未发送等的总和，但不是每个msisdn。结果如下：

当我在msisdn条款中添加GROUP BY时，我不会按照我需要的方式得到结果。

我需要这样：

请帮助我编写优化查询，以便在过去8周内为每个msisdn获取这些结果，因为我被卡住了。

Answer 1

WEEK(...)在今年的第一天附近有问题。相反，您可以使用TO_DAYS：

WHERE datetime > CURDATE() - INTERVAL 8 WEEK   -- for the last 8 weeks
GROUP BY MOD(TO_DAYS(datetime), 7)   -- group by week

这很简单，但它有一个错误。它只适用于今天是＆＃34;周＆＃34;的最后一天。如果日期％7落在所需的星期几。

WHERE datetime > CURDATE() - INTERVAL 9 WEEK   -- for the last 8 weeks
GROUP BY MOD(TO_DAYS(datetime) - 3, 7)   -- group by week

第一次修复错误 - 9周间隔将包括当前的部分周和8周前的部分周。 ＆＃34; - 3＆＃34; （或任何数字的作品）将调整你的＆＃34;周＆＃34;从星期一或星期日或其他什么开始。

如果该列只有0或1个值，则

SUM(IF (sent_messages.delivered = 1,1,0 ))可以缩短为SUM(delivered = 1)甚至SUM(delivered)。