考虑以下表格:
users messages
------------------- ----------------------
user_id avg_quality msg_id user_id quality
------------------- ----------------------
1 1 1 1
2 2 1 0
3 3 1 0
4 1 1
5 1 1
6 2 0
7 2 0
8 3 1
messages.quality
可以是0
或1
。我需要计算每个用户的平均邮件质量并相应地更新users.avg_quality
。因此,所需的输出将被修改为users
表,如下所示:
users
-------------------
user_id avg_quality <-- DECIMAL (8,2)
-------------------
1 0.60 <-- (3x1 + 2x0) / 5
2 0.00 <-- (2x0) / 2
3 1.00 <-- (1x1) / 1
我已经开始了这样的查询,我知道语法不正确但没有更好的主意。你呢?
UPDATE messages m, users u
SET avg_quality = (SELECT COUNT(m.msg_id) / SUM(m.quality))
WHERE m.user_id = u.user_id
答案 0 :(得分:1)
查看平均功能:
http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html#function_avg
您的选择应该是这样的:
select user_id, AVG(quality) from messages group by user_id
如果您从一个空的用户表开始,您可以像这样运行一个查询来全部更新:
insert into users (user_id, avg_quality)
select m.user_id, coalesce(AVG(m.quality),0) from messages m group by m.user_id
如果您需要持续的结果,Luc的建议将适合您:
update users u left join (
select m.user_id, AVG(m.quality) as average from messages m group by m.user_id
) as average_result_t on u.user_id = average_result_t.user_id
set u.average = coalesce(average_result_t.average,0)
答案 1 :(得分:1)
这应该有效:
UPDATE users u
INNER JOIN (SELECT a.user_id, AVG(quality) avg_quality
FROM messages a
INNER JOIN users b
ON a.user_id = b.user_id
GROUP BY a.user_id
) tmp
ON u.user_id = tmp.user_id
SET u.avg_quality = tmp.avg_quality;