SELECT
Day,
month,
year,
GROUP_CONCAT(total),
GROUP_CONCAT(SP_ID)
FROM
(
SELECT
DATE_FORMAT(l.act_date, '%d') AS DAY,
DATE_FORMAT(l.act_date, '%M') AS MONTH,
EXTRACT(YEAR FROM l.act_date) AS YEAR,
COUNT(*) as total,l.sp_id
FROM lead_activity2 as l
right outer join salesperson as s on l.sp_id=s.sp_id
WHERE l.act_name='scb'
AND ((l.act_date>='2012-09-07 13:03:27' )
AND (l.act_date<= '2012-11-07 13:03:27'))
GROUP BY MONTH, YEAR, DAY, l.sp_id
ORDER BY YEAR DESC, MONTH DESC, DAY DESC, l.sp_id DESC
) t GROUP BY day, month, year
http://sqlfiddle.com/#!2/1514d/3 - 您可以查看方案和查询
我想得到的是
18 | october | 2012 | 0,0,1,1 | 6,5,4,3
spid 6和spid 5没有18月10日的数据,但仍然应该显示尝试做正确的连接和右外连接似乎都不起作用...
答案 0 :(得分:1)
像这样使用GROUP_CONCAT
:
SELECT
Day,
month,
year,
GROUP_CONCAT(total),
GROUP_CONCAT(SP_ID)
FROM
(
SELECT
DATE_FORMAT(l.act_date, '%d') AS DAY,
DATE_FORMAT(l.act_date, '%M') AS MONTH,
EXTRACT(YEAR FROM l.act_date) AS YEAR,
COUNT(*) as total,l.sp_id
FROM lead_activity2 as l
WHERE l.act_name='scb'
AND ((l.act_date>='2012-09-07 13:03:27' )
AND (l.act_date<= '2012-11-07 13:03:27'))
GROUP BY MONTH, YEAR, DAY, l.sp_id
ORDER BY YEAR DESC, MONTH DESC, DAY DESC, l.sp_id DESC
) t GROUP BY day, month, year
更新:是的,您可以执行此操作,但使用LEFT JOIN
包含不匹配的sp_id
。这些不匹配的ID将使用NULL
使用IFNULL
的值来显示零,如下所示:
SELECT
Day,
month,
year,
GROUP_CONCAT(total) Total,
GROUP_CONCAT(SP_ID) 'List of sp_ids'
FROM
(
SELECT
DATE_FORMAT(l.act_date, '%d') AS DAY,
DATE_FORMAT(l.act_date, '%M') AS MONTH,
EXTRACT(YEAR FROM l.act_date) AS YEAR,
COUNT(*) as total,
IFNULL(s.sp_id , 0) sp_id
FROM lead_activity2 as l
LEFT JOIN salesperson s ON l.sp_id = s.sp_id
WHERE l.act_name='scb'
AND ((l.act_date>='2012-09-07 13:03:27' )
AND (l.act_date<= '2012-11-07 13:03:27'))
GROUP BY MONTH, YEAR,DAY,s.sp_id
) t
ORDER BY YEAR DESC,
MONTH DESC,
DAY DESC,
sp_id DESC