获取一周一周的数据

Question

我有一张表如下

CREATE TABLE messages  ( 
id          int(10) UNSIGNED AUTO_INCREMENT NOT NULL,
messagetext text NOT NULL,
created_at  timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
PRIMARY KEY(id)

）

我想要做的是获得两周的数据，当前周和前一周，按天排序。例如，在星期日，星期一，星期二，星期三......本周和上周创建了多少条消息。

我的查询如下

SELECT
    COUNT(id) AS 'Count' 
FROM
    'messages'
WHERE
    (WEEK(created_at) = WEEK(NOW())) 
GROUP BY
    WEEKDAY(created_at) 
ORDER BY
    week(created_at) ASC,
    weekday(created_at) ASC

但我不知道如何整理它。例如，上面的查询将列出我今天或昨天获得了多少，但这不是我想要的。我想要的是明天和明智的计算日。

这是我想要的

如果我对自己面临的情况有足够的清楚，请告诉我。

更新 有关更多信息，请参阅下面的Juan的问题

如果当前天数少于14天，那么在这种情况下，脚本应该返回上周的天数，而不是前一周的前一天。以下是胡安的解释（想象今天是星期五。如果你算上14天回来，你星期一星期五到星期五。星期一星期一到星期日...并且上周五星期五到星期日。并且不确定这是否是你的结果想要）

问候

加甘

Answer 1

我要发一个大猜测。但你尝试过这样的事吗？

SELECT
    week(created_at) week_id,
    weekday(created_at) weekday_id,
    COUNT(id) AS 'Count' 
FROM
    'messages'
WHERE
      week_id = WEEK( NOW() ) 
  OR  week_id = WEEK( DATE_SUB(NOW(), INTERVAL 7 DAY) )
GROUP BY
    week_id, weekday_id 
ORDER BY
    week_id, weekday_id

Answer 2

使用条件SUM

SELECT week(created_at) week_id,
       SUM( IF(weekday(created_at) = 0, 1, 0) ) as monday,
       SUM( IF(weekday(created_at) = 1, 1, 0) ) as tuesday,
       SUM( IF(weekday(created_at) = 2, 1, 0) ) as wednesday,
       SUM( IF(weekday(created_at) = 3, 1, 0) ) as thursday,
       SUM( IF(weekday(created_at) = 4, 1, 0) ) as friday,
       SUM( IF(weekday(created_at) = 5, 1, 0) ) as saturday,
       SUM( IF(weekday(created_at) = 6, 1, 0) ) as sunday
FROM  'messages'
WHERE created_at >= DATE_SUB(NOW(), INTERVAL 14 DAY)
  AND (      week_id = WEEK( NOW() ) 
         OR  week_id = WEEK( DATE_SUB(NOW(), INTERVAL 7 DAY) 
      )
GROUP BY
    week_id
ORDER BY
    week_id