MySQL计算列值实例并按列值排序

Question

我有一个相当高级的查询，我需要在一个表上进行保存客户发送的故障单，收集有关故障单的统计信息以及按各种计数和/或列值进行排序和排序。

门票表：

+---------------------------+       
| product_id | issue        |
+------------|--------------+
| 1          | missing-part |
| 1          | missing-part |
| 1          | broken-part  |
| 2          | broken-part  |
| 2          | broken-part  |
| 2          | missing-part |
| 2          | missing-part |
| 2          | missing-part |
+---------------------------+

首先，我需要计算每种产品的门票总数。 在这种情况下， product_id 1有3张票，而product_id 2有5张票。 然后，我需要获取issue列的值以及每个值的出现次数。 所以在这一点上，我期望这样的事情：

[
  'product_id' => 1,
  'issues'     => 3
  'issue_nums' => [
    'missing-part' => 2,
    'broken-part'  => 1
  ]
],
[
  'product_id' => 2,
  'issues'     => 5,
  'issue_nums' => [
    'missing-part' => 3,
    'broken-part'  => 2
  ]
]

然后，我需要能够按每个问题的出现次数进行排序，例如： ORDER BY COUNT(missing-part)或COUNT(broken-part) 。 在其他情况下，我只需要能够按总问题数而不是问题值本身来订购产品（因此，按COUNT(issues)订购）即可。

谢谢！

Answer 1

yu可以使用subselect来计数product_id的出现，并使用main select来进行issuse的计数

  select t1.product_id, t2.product_count, t1.issue, count(*) count_issue
  from  Tickets t1
  inner join  (
  select product_id, count(*)  product_count
  from Tickets 
  group by product_id) t2 on t2.product_id = t1.product_id
  group by t1.product_id, t2.product_count, t1.issue
  order by  count_issue

Answer 2

您可以非常接近两个聚合级别：

select t.product_id, sum(cnt) as total,
       group_concat(issue, ':', cnt separator ',') as issues
from (select t.product_id, t.issue, count(*) as cnt
      from tickets t
      group by t.product_id, t.issue
     ) tp
group by t.product_id;