Question

我有一个名为'services'的MySql表，其结构为：机构，角色

    Institution     |   Role
    Human Ressource |   Responsable
    Human Ressource |   Responsable
    Human Ressource |   Agent
    Human Ressource |   Agent
    Human Ressource |   Chief
    IT Service      |   Responsable
    IT Service      |   Agent
    Client Relate   |   Responsable 
    Client Relate   |   Chief
    Control Mgm     |   Responsable
    Control Mgm     |   Agent
    Control Mgm     |   Agent

和一个查询以计算每个机构的角色数量，例如：

    SELECT a.institution, Responsable, Agent, Chief
    FROM
    (
      SELECT a.institution
      FROM services a
      WHERE a.month = '2019-04'
      GROUP BY a.institution
    ) a,
    (
      SELECT b.institution, NULLIF(count(b.role) ,0) AS Responsable
      FROM services b
      WHERE
        b.month = '2019-04' AND
        b.role = 'Responsable'
      GROUP BY b.institution
    ) b,
    (
      SELECT c.institution, NULLIF(count(c.role), 0) AS Agent
      FROM services c
      WHERE
        c.month = '2019-04' AND
        c.role = 'Agent'
      GROUP BY c.institution
    ) c,
    (
      SELECT d.institution, NULLIF(count(d.role), 0) AS Chief
      FROM services d
      WHERE
        d.month = '2019-04' AND
        d.role = 'Chief'
      GROUP BY d.institution
    ) d
    WHERE
      a.institution = b.institution AND
      a.institution = d.institution
    GROUP BY 1
    ORDER BY a.institution ASC

我想在一个sql语句中达到的结果是：

    Institution     |   Responsable |   Agent   |   Chief   
    Human Ressource |   2           |   2       |   1
    IT Service      |   1           |   1       |   0
    Client Relate   |   1           |   0       |   1
    Control Mgm     |   1           |   2       |   0

但是我得到一个非常奇怪的结果集

    institution     |   Responsable |   Agent   |   Chief
    Client Relate   |   1           |   2       |   1
    Human Ressource |   2           |   2       |   1

我在这里创建了sqlfiddle

任何想法都会受到高度赞赏。 Tnx

Answer 1

使用条件聚合

SELECT a.institution,
       coalesce(count(case when role = 'Responsable' then 1 end),0) AS Responsable,
       coalesce(count(case when role = 'Agent' then 1 end),0) AS Agent,
       coalesce(count(case when role = 'Chief' then 1 end),0) AS Chief
FROM services a
WHERE a.month = '2019-04'
GROUP By a.institution

Answer 2

您可以使用CASE语句，然后汇总结果来使用它：

SELECT  institution,
        SUM(Responsable) AS Responsable,
        SUM(Agent) AS Agent,
        SUM(Chief) AS Chief
  FROM  (
        SELECT  institution,
                CASE
                  WHEN role = 'Responsable' THEN 1
                  ELSE 0
                END AS Responsable,
                CASE
                  WHEN role = 'Agent' THEN 1
                  ELSE 0
                END AS Agent,
                CASE
                  WHEN role = 'Chief' THEN 1
                  ELSE 0
                END AS Chief
          FROM services
          WHERE `month` = '2019-04'
        ) a
  GROUP BY institution
  ORDER BY institution ASC;

这将创建一个结果集，该结果集的每一行的相应列中的数字为1，然后根据该机构将每一行的计数按institution和SUM分组。 / p>

结果：

Client Relate   1   0   1
Control Mgm     1   2   0
Human Ressource 2   2   1
IT Service      1   1   0

显示此内容的SQL Fiddle。

Answer 3

感谢您提供表格架构，示例数据和所需的输出。
试试这个。

SELECT institution,
       Sum(CASE WHEN role = 'Responsable' THEN 1 ELSE 0 end) AS Responsable,
       Sum(CASE WHEN role = 'Agent' THEN 1 ELSE 0 end) AS Agent,
       Sum(CASE WHEN role = 'Chief' THEN 1 ELSE 0 end) AS Chief
FROM   services
GROUP  BY institution

如果需要，还可以添加where子句。
示例：WHERE month = '2019-04'

输出

+------------------+--------------+--------+-------+
|   institution    | Responsable  | Agent  | Chief |
+------------------+--------------+--------+-------+
| Client Relate    |           1  |     0  |     1 |
| Control Mgm      |           1  |     2  |     0 |
| Human Ressource  |           2  |     2  |     1 |
| IT Service       |           1  |     1  |     0 |
+------------------+--------------+--------+-------+

SQL小提琴：http://sqlfiddle.com/#!9/7163f5/16/0

获取从同一表合并的计数数字

3 个答案: