返回按周分组的行数 MySQL

Question

我试图从数据按周分组的 MySQL 表中获取行数。

到目前为止，我的查询是：

 "SELECT count(*) as tweets, twitTimeExtracted as date 
  FROM scene.twitData 
  group by week(twitTimeExtracted)"

此查询返回以下数据：

如您所见，周数不正确，我期待从 1 月 7 日星期一（7、14、21、28、4、11 等）开始到本周的每周数据。

我还尝试了原始查询的修改版本：

SELECT count(*) as tweets, twitTimeExtracted as date 
FROM scene.twitData 
WHERE date(twitTimeExtracted) 
   BETWEEN '2013-01-07' and '2013-03-11' 
group by week(twitTimeExtracted)

这将返回与第一个查询类似的结果。

也许在几行数据上存储在 DATETIME: twitTimeExtracted 列中的某些数据存在不一致？ 我真的不知道我对 MySQL 不是很有经验。

任何帮助将不胜感激。

谢谢

Answer 1

这会将datetime值转换为一周中适当的星期一

select count(*) as tweets,
       str_to_date(concat(yearweek(twitTimeExtracted), ' monday'), '%X%V %W') as `date`
from twitData 
group by yearweek(twitTimeExtracted)

yearweek返回周和年。 str_to_date与字符串monday str_to_date为您提供星期一的datetime值。

如果您的一周从Monday开始，请改用yearweek(twitTimeExtracted, 1) 。

Answer 2

始终获取一周中的星期一的一种选择是使用 adddate：

SELECT count(*) as tweets, adddate(twitTimeExtracted, INTERVAL 2-DAYOFWEEK(twitTimeExtracted) DAY)
FROM twitData 
WHERE date(twitTimeExtracted) 
   BETWEEN '2013-01-07' and '2013-03-11' 
GROUP BY adddate(twitTimeExtracted, INTERVAL 2-DAYOFWEEK(twitTimeExtracted) DAY)

SQL 小提琴演示

Answer 3

试试这个：

 SELECT count(*) as tweets, 
     date_add('7 Jan 2012', day, datediff('7 Jan 2012', twitTimeExtracted)) % 7   
 FROM scene.twitData 
 group by date_add('7 Jan 2012', day, datediff('7 Jan 2012', twitTimeExtracted)) % 7

Answer 4

替代解决方案：

SELECT DATE(DATE_SUB(yourDate, INTERVAL WEEKDAY(yourDate) DAY));

例如：

SELECT DATE(DATE_SUB(DATE("2016-04-11"), INTERVAL WEEKDAY(DATE("2016-04-11")) DAY));

Answer 5

你也可以试试这个。

select 
DATE_ADD(DATE(date), INTERVAL (7 - DAYOFWEEK(date)) DAY) as weekend,
count(tweets)
from scene.twitData 
  group by weekend;