将组中的最大值附加到每一行结果？答案

【问题标题】：Attach max from group to every row of results ?将组中的最大值附加到每一行结果？
【发布时间】：2015-12-18 10:41:05
【问题描述】：

我有类似如下的数据库结构：应用服务表：

+-----+--------------+----------+---------------+
|  id | name         | status   | application_id
|  1  | Service 1    |  1       |     24
|  2  | Service 2    |  2       |     24
|  3  | Service 3    |  3       |     25
+-----+--------------+----------+----------------

还有其他带有状态定义的表：客户状态

+------------+--------------+----------+-----------+
|  status_id | name         | level    | is_closed
|  1         | Status 1     |  2       |     1
|  2         | Status 2     |  1       |     0
|  3         | Status 3     |  3       |     1     
+------------+--------------+----------+----------

ApplicationServices 每一行的状态计算为按 application_id 分组的记录中的最高级别状态。

因此，从 ApplicationServices 中获取所有具有状态的记录将导致如下结果：

+-----+--------------+----------+----------------+-------------+-----------
|  id | name         | status   | application_id | status_name | is_closed
|  1  | Service 1    |  1       |     24         |  Status 1   |    1
|  2  | Service 2    |  2       |     24         |  Status 1   |    1
|  3  | Service 3    |  3       |     25         |  Status 3   |    1
+-----+--------------+----------+----------------+-------------+-----------

有没有一种有效的方法可以将 max(level) 按 application_id 分组的结果附加到结果集的每一行？

【问题讨论】：

标签： mysql sql select group-by greatest-n-per-group

【解决方案1】：

试试这个：

SELECT A.id, A.name, A.status, A.application_id, CS.name AS status_name, CS.is_closed
FROM ApplicationService A
INNER JOIN (SELECT AA.application_id, MAX(CS.level) AS maxLevel
            FROM ApplicationService AA
            INNER JOIN CustomerStatus CS ON AA.status = CS.status_id
            GROUP BY AA.application_id
          ) AS AA ON A.application_id = AA.application_id 
INNER JOIN CustomerStatus CS ON AA.maxLevel = CS.level;

【讨论】：

这正是我所需要的，实际上几乎是我以前的两倍。谢谢！

【解决方案2】：

See working SQLfiddle here.

您可能必须执行子查询。子查询只需将两个表连接在一起，按application_id 分组/折叠，然后获取最大is_closed 值：

SELECT
     t1.status AS `status`,
     t1.application_id AS `app_id`,
     MAX(t2.is_closed) AS `max_is_closed`
   FROM applicationstatus AS t1
  LEFT JOIN customerstatus AS t2 ON
    t1.status = t2.status_id
  GROUP BY t1.application_id

当您合并子查询时，is_closed 的最大值应该可以通过 max_is_closed 别名访问：

SELECT
  t1.id AS id,
  t1.name AS name,
  t1.status AS `status`,
  t1.application_id AS application_id,
  t2.name AS status_name,
  t3.max_is_closed
FROM applicationstatus AS t1
LEFT JOIN customerstatus AS t2 ON
  t1.status = t2.status_id
LEFT JOIN
  (SELECT
     t1.status AS `status`,
     t1.application_id AS `app_id`,
     MAX(t2.is_closed) AS `max_is_closed`
   FROM applicationstatus AS t1
  LEFT JOIN customerstatus AS t2 ON
    t1.status = t2.status_id
  GROUP BY t1.application_id) AS t3 ON
  t1.application_id = t3.app_id

查询的输出：

【讨论】：