Question

我有一张这样的表：

Employee
| id | name | department |
| 01 | Joe  | Network    |
| 02 | Sam  | Quality    |
| 03 | Mark | Network    |
| 04 | Pete | Quality    |

和其他表格一样：

Hours
| id | date    | hours |
| 01 | 1/1/11  | 7     |
| 02 | 1/2/11  | 15    |
| 03 | 1/5/11  | 13    |
| 01 | 1/7/11  | 5     |
| 01 | 1/11/11 | 9     |
| 02 | 1/11/11 | 11    |
| 03 | 1/12/11 | 14    |
| 04 | 1/12/11 | 14    |

我想查询一下：显示每个部门达到最大总小时数的人（从最大值到最小值排序）

| id | Name | Department | totalhours |
| 03 | Mark | Network    | 27         |
| 02 | Sam  | Quality    | 26         |

我目前的代码不起作用，只显示每个人的总小时数：

SELECT e.name, e.department, SUM(h.hours) AS total
FROM employee e JOIN hours h ON e.id = h.id
GROUP BY e.name, e.department
ORDER BY total DESC;

我需要做什么？

我试过这样的事情......

SELECT e.name, e.department, t.total
FROM (
    SELECT e2.department, SUM(h.hours) AS total
    FROM employee e2 JOIN hours h ON e2.id=h.id 
    GROUP BY e2.department, h.hours
    ) t JOIN employee e JOIN hours h ON e.id=h.id ON e.department = t.department AND t.total = h.hours

ORDER BY t.total DESC;

但这显示了疯狂的结果（我认为我的代码很疯狂）

请帮助!! 谢谢！

Answer 1

如果存在平局，这将为部门显示超过1行。可能你想要的是什么。

    SELECT EachEmp.name, EachEmp.department, MAXES.maxTotal AS total
    FROM (SELECT e.name,    -- sum up each employees totals
                 e.department, 
                 SUM(h.hours) AS total
          FROM employee e 
          JOIN hours h ON e.id = h.id
          GROUP BY e.name, e.department) EachEmp
    JOIN
       (SELECT department, max(total) maxTotal  -- get the maximum emp total for this dept
        FROM (   SELECT e.department, SUM(h.hours) AS total
                 FROM employee e 
                 JOIN hours h ON e.id = h.id
                 GROUP BY e.name, e.department) AS TOTAL -- note we are grouping by e.name
        GROUP BY department) AS MAXES
    ON EachEmp.department = MAXES.department
       AND EachEmp.total = MAXES.maxTotal
    ORDER BY MAXES.maxTotal DESC;

Answer 2

在Postgres中，最简单（通常也是最快）的方法是distinct on：

SELECT DISTINCT ON (e.name) e.name, e.department, SUM(h.hours) AS total
FROM employee e JOIN
     hours h
     ON e.id = h.id
GROUP BY e.name, e.department
ORDER BY e.name, total DESC;

SQL希望避免重复的行

3 个答案: