Question

（这是我以前的问题的重写，可能不够清楚）

我有一个MYSQL数据库的查询，如下所示：

SELECT name,
SUM(IF(date_format (date, '%b, %Y')= 'Dec, 2011', 1,0)) AS `month1`, 
SUM(IF(date_format (date, '%b, %Y')= 'Jan, 2012', 1,0)) AS `month2`, 
SUM(IF(date_format (date, '%b, %Y')= 'Feb, 2012', 1,0)) AS `month3`, 
etc...

这给了我一系列结果，如 - month1 = 55，month2 = 70，month3 = 89等

在查询中是一行 -

COUNT(*) AS total FROM table order by total

这实际上给了我一个月1 +月2 +月3 +等

但是，我还需要获得相同月度总数的平均值

所以我需要一个有效的类似

的MySQL函数

AVG (month1, month2, month3 etc)

这将给出平均55,70,89

有人可以帮忙吗？

非常感谢

根据要求，完整的查询 -

SELECT name, 
    SUM(IF(date_format (date, '%b, %Y')= 'Nov, 2011', 1,0))/list*1000 AS `month1`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Dec, 2011', 1,0))/list*1000 AS `month2`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Jan, 2012', 1,0))/list*1000 AS `month3`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Feb, 2012', 1,0))/list*1000 AS `month4`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Mar, 2012', 1,0))/list*1000 AS `month5`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Apr, 2012', 1,0))/list*1000 AS `month6`, 
    SUM(IF(date_format (date, '%b, %Y')= 'May, 2012', 1,0))/list*1000 AS `month7`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Jun, 2012', 1,0))/list*1000 AS `month8`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Jul, 2012', 1,0))/list*1000 AS `month9`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Aug, 2012', 1,0))/list*1000 AS `month10`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Sep, 2012', 1,0))/list*1000 AS `month11`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Oct, 2012', 1,0))/list*1000 AS `month12`, 
    COUNT(*) AS total 
FROM table 
group by name 
order by total

Answer 1

在您的情况下，您可以使用子查询 -

SELECT name,
  `month1`, `month2`, `month3`
  total,
  (`month1` + `month2` + `month3`) / 3 AS `avg`
FROM
  (SELECT name, 
    SUM(IF(date_format (date, '%b, %Y')= 'Nov, 2011', 1,0))/list*1000 AS `month1`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Dec, 2011', 1,0))/list*1000 AS `month2`, 
    SUM(IF(date_format (date, '%b, %Y')= 'Jan, 2012', 1,0))/list*1000 AS `month3`, 
    COUNT(*) AS total 
  FROM table 
  GROUP BY name 
  ORDER BY total
  ) t

但我建议你使用这样的东西 -

SELECT month, AVG(cnt) cnt FROM
  (SELECT MONTH(DATE) month, COUNT(*) cnt FROM table1 GROUP BY month) t
GROUP BY month WITH ROLLUP

...你只应该加上年度支持。

Answer 2

您只需使用

即可

SELECT name, month1,month2,...., AVG(month1,month2,...,month12) as Average FROM
(
SELECT name,
SUM(IF(date_format (date, '%b, %Y')= 'Dec, 2011', 1,0)) AS `month1`, 
SUM(IF(date_format (date, '%b, %Y')= 'Jan, 2012', 1,0)) AS `month2`, 
SUM(IF(date_format (date, '%b, %Y')= 'Feb, 2012', 1,0)) AS `month3`, 
etc...
) as t

获取MySQL查询和SUMS中的平均变量

2 个答案: