平均使用 MySQL 的天数答案

【问题标题】：Average number of days using MySQL平均使用 MySQL 的天数
【发布时间】：2012-07-20 03:27:16
【问题描述】：

我试图获取用户关闭工单到打开“下一个”工单之间的平均天数。

下面是我的 MySQL 表示例。您会注意到 Mike 创建了 3 张工单，间隔 9 天（7 月 1 日-7 月 10 日）和 5 天（7 月 10 日-7 月 15 日），平均为 7 天。我似乎无法弄清楚如何寻找最后解决的日期，有人有什么想法吗？这是我目前所拥有的：

SELECT 
  Name, 
  Created, 
  Resolved, 
  avg(datediff("Last Ticket Resolved", created) AS last_tket_open 
FROM 
  MyTable 
WHERE 
  Name='Mike'


Name   Created  Resolved
----   -------  --------
Mike   July  1   July  1
Jill   July  2   July  3    
Mike   July 10   July 10
Harry  July 11   July 11
Mike   July 15   July 15

【问题讨论】：

标签： mysql average

【解决方案1】：

SELECT a.name, 
       AVG(DATEDIFF(a.resolved, b.created)) AS avgdays
FROM
(
   SELECT name, resolved, @val1:=@val1+1 AS rn
   FROM tbl
   CROSS JOIN (SELECT @val1:=0) val1_init
   WHERE name = 'Mike'
   ORDER BY resolved
) a
INNER JOIN
(
   SELECT created, @val2:=@val2+1 AS rn
   FROM tbl
   CROSS JOIN (SELECT @val2:=1) val2_init
   WHERE name = 'Mike'
   ORDER BY resolved
) b ON a.rn = b.rn

SQLFiddle Demo

【讨论】：

谢谢，这正是我想要的。也感谢 SQLFiddle 演示链接，如果我以后有问题，我会使用它。

【解决方案2】：

select 
  Name,
  AVG( datediff( next_created, resolve )) AS avg_time
from (select
  t1.Name,
  t1.Created,
  t1.Resolved,
  min(t2.Created) as next_created
from mytable t1
join myTable t2 on t1.Name = t2.Name and t2.Created > t1.Resolved
group by 1,2,3) x
group by 1;

请注意，此查询不使用相关子查询（由于编码方式依赖于行中的值而对每一行执行的查询），因此它应该执行得相当好。

【讨论】：

【解决方案3】：

试试这个：

SELECT
  t1.Name,
  AVG( datediff( t2.Created, t1.Resolved )) AS avg_days_between_tickets
FROM
  MyTable t1,
  MyTable t2
WHERE
      t2.Name = t1.Name
  AND t2.Created = ( SELECT MIN( Created ) 
                     FROM MyTable 
                     WHERE Name = t1.Name 
                       AND Created > t1.Resolved )
GROUP BY
  t1.Name

【讨论】：