Question

我有这个查询，从03-03-2017到现在输出（排序）每天的日期和每天的计数。有没有办法按周分组（7天）？

query.sql的：

SELECT SUM(a.ctr)
,a.datecreated
FROM 
(
    SELECT COUNT(*) as ctr, date_format(created, '%d/%m/%y ') as datecreated 
    FROM mimesi_indexer.meta_served_clips
    GROUP BY date(datecreated)
    UNION ALL 
    select 0 as ctr, date_format(selected_date,  '%d/%m/%y ') as datecreated 
       from 
     (select adddate('1970-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) selected_date from
     (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
     (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
     (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
     (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
     (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
    where selected_date 
    between (SELECT MIN(created) FROM mimesi_indexer.meta_served_clips)
        and (SELECT MAX(created) FROM mimesi_indexer.meta_served_clips)
) a
group by date(a.datecreated)
order by month(a.datecreated), date(a.datecreated)

结果（部分内容）：

'279',  '03/03/17'
'40',   '04/03/17'
'41',   '05/03/17'
'1223', '06/03/17'
'493',  '07/03/17'
'580',  '08/03/17'
'829',  '09/03/17'
'936',  '10/03/17'
'638',  '11/03/17'
'345',  '12/03/17'
'876',  '13/03/17'
'1583', '14/03/17'
'1566', '15/03/17'
'1772', '16/03/17'

想要的结果：

'3485',  '03/03/17'
'7716',  '10/03/17'

编辑结果按周使用分组（a.datecreated）：

'38657', '03/03/17 '
'4773', '07/03/17 '
'28529', '01/04/17 '
'14191', '05/04/17 '
'34223', '01/05/17 '
'43093', '04/05/17 '
'82515', '01/06/17 '
'1804', '12/06/17 '
'23513', '01/07/17 '
'9287', '05/07/17 '

P.S： 如果一周（7天）可以从2017年3月5日（05-03-2017）开始，那将是完美的，因为这是星期一，第3和第4将保留为单个数字

Answer 1

Activity

使用此行SELECT SUM(a.ctr) ,a.datecreated FROM ( SELECT COUNT(*) as ctr, date_format(created, '%d/%m/%y ') as datecreated FROM mimesi_indexer.meta_served_clips GROUP BY date(datecreated) UNION ALL select 0 as ctr, date_format(selected_date, '%d/%m/%y ') as datecreated from (select adddate('1970-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) selected_date from (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0, (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1, (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2, (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3, (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v where selected_date between (SELECT MIN(created) FROM mimesi_indexer.meta_served_clips) and (SELECT MAX(created) FROM mimesi_indexer.meta_served_clips) ) a group by week(STR_TO_DATE(a.datecreated,'%d/%m/%y')) order by month(a.datecreated), date(a.datecreated)。

您正在将字符串字段转换为日期，然后将该日期转换为星期。

您可以尝试以上查询。

Answer 2

应该使用：

GROUP BY WEEK(a.datecreated, 7)

或者查看documentation of function WEEK()获取第二个参数的值。

按周（7天）查询的组结果 - MySql

2 个答案: