Question

在有人评论之前，我没有用逗号分隔值设计这个数据库:) 我花时间试图找到答案，但我能找到的只是GROUP_CONCAT（），这似乎与我想要的相反。

我想在逗号分隔值字段中为每个值分组。

SELECT round(avg(DATEDIFF( dateClosed , dateAded ) * 1.0), 2) AS avg, department
FROM tickets GROUP BY assignedto

assignedto字段是逗号分隔值字段

row1 54,69,555
row2 54,75,555
row3 75,555

DESIRED OUTPUT：分配给字段中每个值的平均舍入数字。

编辑 - 尝试将其带到下一级：我想要包含故障单答案表以获取该故障单的第一个响应，使用其日期时间字段计算每个用户的平均响应时间。

SELECT a.id as theuser, round(avg(DATEDIFF( ta.dateAded , t.dateAded ) * 1.0), 2) as avg
FROM tickets t join
     mdl_user a
     on find_in_set(a.id, t.assignedto) > 0
INNER JOIN (SELECT MIN(ta.dateAded) as started FROM ticketanswer GROUP BY ta.ticketId) ta ON t.id = ta.ticketId
GROUP BY a.id ORDER BY avg ASC

Answer 1

呸。假设您知道最大分配数，则可以执行此操作。这是一种方法：

select substring_index(substring_index(assignedto, ',', n.n), ',', -1) as assignedto,
       round(avg(DATEDIFF( dateClosed , dateAded ) * 1.0), 2) as avg
from tickets t join
     (select 1 as n union all select 2 union all select 3)
     on length(assignedto) - length(replace(assignedto, ',', '')) < n.n
group by substring_index(substring_index(assignedto, ',', n.n), ',', -1);

或者，如果您有一个已分配值的列表，比如在AssignedTo表中，则更简单：

select a.assignedto, round(avg(DATEDIFF( dateClosed , dateAded ) * 1.0), 2) as avg
from tickets t join
     assignedto a
     on find_in_set(a.assignedto, t.assignedto) > 0
group by a.assignedto;

对不起，您必须处理这种格式错误的数据库结构。

MySQL GROUP BY每个逗号分隔值

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