Question

我当前的查询：

    Select `userSubUnitsID`, `userSubUnits`.`userID`, CONCAT(`u`.`fname`,' ',`u`.`lname`) AS `full`, `pID`, SUM(`quantComp`) AS `total`
    From `userSubUnits`
    JOIN `u` ON `u`.`userID` = `userSubUnits`.`userID`
    GROUP BY `userID`

我需要的是约会。所以看起来会更像这样。

    +-----+------------+----------+-------------+
    |user |current week|prev week |two weeks ago|
    +-----+------------+----------+-------------+
    |John |         564|       354|          687|
    +-----+------------+----------+-------------+
    |Paul |         451|       328|          845|
    +-----+------------+----------+-------------+
    etc

单个查询是否可以实现？我知道如何限制本周，但不知道如何在前几周添加。

仅供参考，周末在周日午夜结束。提前致谢

Answer 1

SELECT user, SUM(WEEKOFYEAR(datefield) = WEEKOFYEAR(now()) AS current_week,
    SUM(WEEKOFYEAR(datefield) = WEEKOFYEAR(now() - INTERVAL 1 WEEK)) AS last_week,
    SUM(WEEKOFYEAR(datefield) = WEEKOFYEAR(now() - INTERVAL 2 WEEK)) AS two_weeks_ago
FROM ...
GROUP BY WEEKOFYEAR(now())

丑陋，但应该做的伎俩。请注意，如果您的日期跨越一年边界，则此操作将失败。为了解决这个问题，你必须添加一些额外的基于年的逻辑。

同样，你最好还是做这种转型的客户端。扩展是丑陋的，如果你需要开始做超过几周的事情，很快就会变得非常低效，例如：

SELECT user, YEAR(datefield), WEEKOFYEAR(datefield), COUNT(*)
FROM ...
GROUP BY YEAR(datefield), WEEKOFYEAR(datefield)

MySQL有几个日期范围

1 个答案: