mysql - 对于员工表中的每个员工，查找层次结构中直接和间接老板的数量

Question

我得到了一个员工表，看起来像这样:

创建示例输入的查询。

CREATE TABLE employee(
    empId INTEGER,
    empName VARCHAR(20),
    mgrId INTEGER,
    salary DECIMAL(12,2)
);

INSERT INTO employee VALUES
(1, 'A', 2, 100),
(2, 'B', 4, 150),
(3, 'C', 4, 165),
(4, 'D', 7, 200),
(5, 'E', 6, 210),
(6, 'F', 7, 250),
(7, 'G', 7, 300),
(8, 'H', 6, 170);

链接到 SQL Fiddle:http://sqlfiddle.com/#!9/cd4be8

此示例数据会产生此层次结构。

每个员工都有一个直接上司。此外，所有比特定员工高级的员工都被称为他们的间接老板。 在给定的样本中，A的直接老板是B，C、D、E、F、G和H是A的间接老板。我需要找到两件事。

1. 对于每个员工，查找比该员工高级的所有员工的数量(直接上司 + 间接上司的数量)。

预期输出:

+---------+-------+
| empName | total |
+---------+-------+
| A       |     7 |
| B       |     3 |
| C       |     3 |
| D       |     1 |
| E       |     3 |
| F       |     1 |
| G       |     0 |
| H       |     3 |
+---------+-------+

对于每个员工，找到间接老板，使间接老板的工资至少是给定员工工资的两倍，但在间接老板中最低。

预期输出:

+---------+------+
| empName | mgr  |
+---------+------+
| A       | D    |
| B       | G    |
| C       | NULL |
| D       | NULL |
| E       | NULL |
| F       | NULL |
| G       | NULL |
| H       | NULL |
+---------+------+

解释:对于员工A(工资=100)，工资至少双倍的间接老板是D(工资=200)、F(工资=210)和G(工资=300)。但由于 D 的工资是 D、F 和 G 中最低的，所以结果是 D。

查找员工的下属人数很容易，但反之则非常棘手。任何帮助/提示将不胜感激。

Answer 1

我希望您会发现以下答案对您有所帮助，直到出现更优化的解决方案。

第一步创建一个新 View ，记录员工的级别。在这种情况下，名为 G 的“老板”的值最小为 1。而每个级别的员工的排名都在增加，其中 A 的最高值为 4。该 View 是使用分层查询构建的 here。 .

创建 View 后，通过对所有较高级别的员工求和来回答第一个问题。这是针对每个排名完成的，并且连接将信息带到最终 View 。

第二个问题以更暴力的方式回答。执行自连接以产生员工和老板的所有可行组合。之后，将生成具有最低老板工资的行。

--Creating a view with the level of the diagram
CREATE VIEW MyCTEView 
AS 
    WITH my_anchor AS (
        SELECT boss.[empId], boss.[empName], boss.[mgrId], 1 AS EmpLevel
        FROM [MySchema].[dbo].[employee] AS boss
        WHERE boss.[mgrId]=boss.[empId]

        UNION ALL

        SELECT Emp.[empId], Emp.[empName], Emp.[mgrId], EL.EmpLevel+1
        FROM [MySchema].[dbo].[employee] AS Emp
        INNER JOIN my_anchor as EL
        ON Emp.[mgrId] = EL.[empId]
        WHERE Emp.[mgrId]<>Emp.[empId]
        )
    SELECT * FROM my_anchor;


--Answer to the first question
SELECT A.[empName]
      ,temp.direct_and_indirect-1 AS your_total
FROM [MySchema].[dbo].[MyCTEView] AS A
LEFT JOIN (
SELECT [EmpLevel],SUM(direct_subortinates) OVER(ORDER BY [EmpLevel]) AS direct_and_indirect
FROM (SELECT COUNT([mgrId]) AS direct_subortinates,[EmpLevel]
            FROM [MySchema].[dbo].[MyCTEView] GROUP BY [EmpLevel])T) AS Temp
ON Temp.[EmpLevel]=A.[EmpLevel]
ORDER BY A.[empName]


--Answer to the second question. Creating a CTE with all the viable combinations of employee and manager based on criteria.
--Displaying the information for the minimum
WITH cte AS (
SELECT A.[empId] as emId
      ,A.[empName] as emName
      ,A.[salary] as emsalary
      ,A.[EmpLevel] as emLevel
      ,B.[empId] as bossId
      ,B.[empName] as bossName
      ,B.[salary] as bosssalary
      ,B.[EmpLevel] as bossLevel
  FROM [MySchema].[dbo].[MyCTEView] AS A
  INNER JOIN 
  [MySchema].[dbo].[MyCTEView] AS B
  ON A.empId<>B.empId AND A.[EmpLevel]>B.[EmpLevel] AND B.[salary]>=2*A.[salary]
  )
SELECT tb1.emName, tb1.bossName 
FROM cte AS tb1
  INNER JOIN
  (
    SELECT emName, MIN(bosssalary) MinSalary
    FROM cte
    GROUP BY emName
  )tb2
  ON  tb1.emName=tb2.emName
  WHERE tb1.bosssalary=tb2.MinSalary