返回按周MySQL分组的行数

Question

我试图从MySQL表中获取行数，其中数据按WEEK分组。

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（7,14,21,28,4,11等等）开始每周的数据并持续到本周

我还尝试了orignal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)

返回与第一个查询类似的结果。

可能存在与DATETIME中存储的一些数据不一致：twitTimeExtracted列中的几行数据？我真的不知道我对MySQL不是很有经验。

真的很感激任何帮助。

由于

Answer 1

这会将datetime值转换为适当的星期一

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。与字符串monday一起，str_to_date为您提供星期一的datetime值。

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

Answer 2

始终获得星期一的一个选择是使用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 Fiddle Demo

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

您也可以尝试这个。

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