sql 自我参考得到平均工资 table
sql get average salary in self reference table
我看过几个关于如何递归查询 self-referencing table 的 questions/answers，但我正在努力应用我找到的答案以汇总到每个 parent、grandparent 等，无论项目在层次结构中的位置如何。
需要获取每个部门（包括层级）的平均工资。
意思是部门应该包括每个人的平均工资sub-department等等。
我有下一个数据库模式：
CREATE TABLE Employee
(
Id INT NOT NULL ,
Name VARCHAR(200) NOT NULL ,
Department_Id INT NOT NULL ,
PRIMARY KEY ( Id )
);
CREATE TABLE Department
(
Id INT NOT NULL ,
DepartmentName VARCHAR(200) NOT NULL ,
Parent_Id INT ,
PRIMARY KEY ( Id )
);
CREATE TABLE Salary
(
Id INT NOT NULL ,
Date DATETIME NOT NULL ,
Amount INT NOT NULL ,
Employee_Id INT NOT NULL ,
PRIMARY KEY ( Id )
);
我试过类似的东西，但它只包含层次结构的第一层。
SELECT d.Id ,
d.DepartmentName ,
( SELECT AVG(s.Amount)
FROM dbo.Department dd
LEFT JOIN dbo.Department sdd ON dd.Id = sdd.Parent_Id
JOIN dbo.Employee e ON e.Department_Id = sdd.Id
OR e.Department_Id = dd.Id
JOIN dbo.Salary s ON s.Employee_Id = e.Id
WHERE dd.Id = d.Id
) AS avg_dep_salary
FROM dbo.Department d
WHERE d.Parent_Id IS NULL;
如何得到各级平均工资？
编辑： 添加了一些插入内容
INSERT INTO Employee
( Id, Name, Department_Id )
VALUES ( 1, 'Peter', 1 ),
( 2, 'Alex', 1 ),
( 3, 'Sam', 2 ),
( 4, 'James', 2 ),
( 5, 'Anna', 3 ),
( 6, 'Susan', 3 ),
( 7, 'Abby', 4 ),
( 8, 'Endy', 4 );
INSERT INTO Department
( Id, DepartmentName, Parent_Id )
VALUES ( 1, 'IT', NULL ),
( 2, 'HR', NULL),
( 3, 'SubIT', 1 ),
( 4, 'SubSubIT', 3 );
INSERT INTO Salary
( Id, Date, Amount, Employee_Id )
VALUES ( 1, '2013-01-09 16:03:50.003', 3000, 1 ),
( 2, '2013-01-11 16:03:50.003', 5000, 2 ),
( 3, '2013-01-09 16:03:50.003', 2000, 3 ),
( 4, '2013-01-11 16:03:50.003', 1000, 4 ),
( 5, '2013-01-09 16:03:50.003', 4000, 5 ),
( 6, '2013-01-11 16:03:50.003', 6000, 6 ),
( 7, '2013-01-09 16:03:50.003', 7000, 7 ),
( 8, '2013-01-13 16:03:50.003', 9000, 8 );
预期结果是：
Department | Average_Salary
__________________________________
IT | ( X1 + X2 + X3 ) / 3
HR | ( Y1 ) / 1
SubIT | ( X2 + X3 ) / 2
SubSubIT | ( X3 ) / 1
其中：
- X1 - IT 部门的平均工资
- X2 - 副IT部门平均工资
- X3 - SubSubIT 部门的平均工资
- Y1 - 平均值
人力资源部工资
这是一种方法
with avg_per_dep as (
select
[Month] = eomonth(s.date), d.Id, d.DepartmentName
, avgDep = avg(s.Amount * 1.0)
from
Salary s
join Employee e on s.Employee_Id = e.Id
join Department d on e.Department_Id = d.Id
group by d.Id, d.DepartmentName, eomonth(s.date)
)
, rcte as (
select
i = Id, Id
, list = cast(',' + cast(Id as varchar(10)) + ',' as varchar(max))
, step = 1
from
Department
union all
select
a.i, b.Id, cast(a.list + cast(b.Id as varchar(10)) + ',' as varchar(max))
, step + 1
from
rcte a
join Department b on a.Id = b.Parent_Id
)
select
d.DepartmentName, c.[Month]
, Average_Salary = avg(c.avgDep)
from
(
select
top 1 with ties i, list
from
rcte
order by row_number() over (partition by i order by step desc)
) t
join avg_per_dep c on t.list like '%,' + cast(c.Id as varchar(10)) + ',%'
join Department d on t.i = d.Id
group by t.i, d.DepartmentName, c.[Month]
输出
DepartmentName [Month] Average_Salary
---------------------------------------------
IT 2013-01-31 5666.666666
HR 2013-01-31 1500.000000
SubIT 2013-01-31 6500.000000
SubSubIT 2013-01-31 8000.000000
想法：
- 计算每个部门的平均工资
- 获取包含递归 CTE 的所有子级的部门列表。
- 合并两个 table 并计算孩子的平均值
示例数据
我添加了几行具有更宽树结构的行。
DECLARE @Employee TABLE
(
Id INT NOT NULL ,
Name VARCHAR(200) NOT NULL ,
Department_Id INT NOT NULL ,
PRIMARY KEY ( Id )
);
DECLARE @Department TABLE
(
Id INT NOT NULL ,
DepartmentName VARCHAR(200) NOT NULL ,
Parent_Id INT ,
PRIMARY KEY ( Id )
);
DECLARE @Salary TABLE
(
Id INT NOT NULL ,
Date DATETIME NOT NULL ,
Amount INT NOT NULL ,
Employee_Id INT NOT NULL ,
PRIMARY KEY ( Id )
);
INSERT INTO @Employee
( Id, Name, Department_Id )
VALUES
( 1, 'Peter', 1 ),
( 2, 'Alex', 1 ),
( 3, 'Sam', 2 ),
( 4, 'James', 2 ),
( 5, 'Anna', 3 ),
( 6, 'Susan', 3 ),
( 7, 'Abby', 4 ),
( 8, 'Endy', 4 ),
(10, 'e_A', 10),
(11, 'e_AB', 11),
(12, 'e_AC', 12),
(13, 'e_AD', 13),
(14, 'e_ACE', 14),
(15, 'e_ACF', 15),
(16, 'e_ACG', 16);
INSERT INTO @Department
( Id, DepartmentName, Parent_Id )
VALUES
( 1, 'IT', NULL ),
( 2, 'HR', NULL),
( 3, 'SubIT', 1 ),
( 4, 'SubSubIT', 3 ),
(10, 'A', NULL ),
(11, 'AB', 10),
(12, 'AC', 10),
(13, 'AD', 10),
(14, 'ACE', 12),
(15, 'ACF', 12),
(16, 'ACG', 12);
INSERT INTO @Salary
( Id, Date, Amount, Employee_Id )
VALUES
( 1, '2013-01-09 16:03:50.003', 3000, 1 ),
( 2, '2013-01-11 16:03:50.003', 5000, 2 ),
( 3, '2013-01-09 16:03:50.003', 2000, 3 ),
( 4, '2013-01-11 16:03:50.003', 1000, 4 ),
( 5, '2013-01-09 16:03:50.003', 4000, 5 ),
( 6, '2013-01-11 16:03:50.003', 6000, 6 ),
( 7, '2013-01-09 16:03:50.003', 7000, 7 ),
( 8, '2013-01-13 16:03:50.003', 9000, 8 ),
(10, '2013-01-13 16:03:50', 100, 10),
(11, '2013-01-13 16:03:50', 100, 11),
(12, '2013-01-13 16:03:50', 100, 12),
(13, '2013-01-13 16:03:50', 100, 13),
(14, '2013-01-13 16:03:50', 100, 14),
(15, '2013-01-13 16:03:50', 100, 15),
(16, '2013-01-13 16:03:50', 100, 16);
查询
WITH
CTE_Departments
AS
(
SELECT
D.Id
,D.Parent_Id
,D.DepartmentName
,SUM(Amount) AS DepartmentAmount
,COUNT(*) AS DepartmentCount
FROM
@Department AS D
INNER JOIN @Employee AS E ON E.Department_Id = D.Id
INNER JOIN @Salary AS S ON S.Employee_Id = E.Id
GROUP BY
D.Id
,D.Parent_Id
,D.DepartmentName
)
,CTE_Recursive
AS
(
SELECT
CTE_Departments.Id AS OriginalID
,CTE_Departments.DepartmentName AS OriginalName
,CTE_Departments.Id
,CTE_Departments.Parent_Id
,CTE_Departments.DepartmentName
,CTE_Departments.DepartmentAmount
,CTE_Departments.DepartmentCount
,1 AS Lvl
FROM CTE_Departments
UNION ALL
SELECT
CTE_Recursive.OriginalID
,CTE_Recursive.OriginalName
,CTE_Departments.Id
,CTE_Departments.Parent_Id
,CTE_Departments.DepartmentName
,CTE_Departments.DepartmentAmount
,CTE_Departments.DepartmentCount
,CTE_Recursive.Lvl + 1 AS Lvl
FROM
CTE_Departments
INNER JOIN CTE_Recursive ON CTE_Recursive.Id = CTE_Departments.Parent_Id
)
SELECT
OriginalID
,OriginalName
,SUM(DepartmentAmount) AS SumAmount
,SUM(DepartmentCount) AS SumCount
,SUM(DepartmentAmount) / SUM(DepartmentCount) AS AvgAmount
FROM CTE_Recursive
GROUP BY
OriginalID
,OriginalName
ORDER BY OriginalID
;
结果
+------------+--------------+-----------+----------+-----------+
| OriginalID | OriginalName | SumAmount | SumCount | AvgAmount |
+------------+--------------+-----------+----------+-----------+
| 1 | IT | 34000 | 6 | 5666 |
| 2 | HR | 3000 | 2 | 1500 |
| 3 | SubIT | 26000 | 4 | 6500 |
| 4 | SubSubIT | 16000 | 2 | 8000 |
| 10 | A | 700 | 7 | 100 |
| 11 | AB | 100 | 1 | 100 |
| 12 | AC | 400 | 4 | 100 |
| 13 | AD | 100 | 1 | 100 |
| 14 | ACE | 100 | 1 | 100 |
| 15 | ACF | 100 | 1 | 100 |
| 16 | ACG | 100 | 1 | 100 |
+------------+--------------+-----------+----------+-----------+
运行 逐步查询，逐个 CTE 以了解其工作原理。
CTE_Departments给出每个部门的总金额和人数。
CTE_Recursive 递归地为每个部门生成子行，同时保持 OriginalID - 递归开始的部门 ID。
最终查询只是按此 OriginalID.
对所有内容进行分组
您也可以使用下面的查询来获得预期的结果
WITH Department_Path
AS (SELECT Id, CAST(CONCAT('@', Id, '@') AS VARCHAR(255)) AS Path
FROM Department
WHERE Parent_Id IS NULL
UNION ALL
SELECT Child.Id, CAST(CONCAT(Parent.Path, Child.Id, '@') AS VARCHAR(255)) AS Path
FROM Department Child
INNER JOIN Department_Path Parent
ON Parent.Id = Child.Parent_Id)
SELECT Department.Id,
Department.DepartmentName,
AVG(Salary.Amount) As Average_Salary,
COUNT(Employee.Id) AS Employee_Count
FROM Department
INNER JOIN Department_Path
ON CHARINDEX(CONCAT('@', Department.Id, '@'), Department_Path.Path) > 0
INNER JOIN Employee
ON Employee.Department_Id = Department_Path.Id
INNER JOIN Salary
ON Salary.Employee_Id = Employee.Id
GROUP BY Department.Id,
Department.DepartmentName;
这个想法是每个员工都属于一个层次结构部门列表。对于每个部门，我们可以检索属于它的所有员工，然后计算平均工资。
我看过几个关于如何递归查询 self-referencing table 的 questions/answers，但我正在努力应用我找到的答案以汇总到每个 parent、grandparent 等，无论项目在层次结构中的位置如何。
需要获取每个部门（包括层级）的平均工资。 意思是部门应该包括每个人的平均工资sub-department等等。
我有下一个数据库模式：
CREATE TABLE Employee
(
Id INT NOT NULL ,
Name VARCHAR(200) NOT NULL ,
Department_Id INT NOT NULL ,
PRIMARY KEY ( Id )
);
CREATE TABLE Department
(
Id INT NOT NULL ,
DepartmentName VARCHAR(200) NOT NULL ,
Parent_Id INT ,
PRIMARY KEY ( Id )
);
CREATE TABLE Salary
(
Id INT NOT NULL ,
Date DATETIME NOT NULL ,
Amount INT NOT NULL ,
Employee_Id INT NOT NULL ,
PRIMARY KEY ( Id )
);
我试过类似的东西，但它只包含层次结构的第一层。
SELECT d.Id ,
d.DepartmentName ,
( SELECT AVG(s.Amount)
FROM dbo.Department dd
LEFT JOIN dbo.Department sdd ON dd.Id = sdd.Parent_Id
JOIN dbo.Employee e ON e.Department_Id = sdd.Id
OR e.Department_Id = dd.Id
JOIN dbo.Salary s ON s.Employee_Id = e.Id
WHERE dd.Id = d.Id
) AS avg_dep_salary
FROM dbo.Department d
WHERE d.Parent_Id IS NULL;
如何得到各级平均工资？
编辑： 添加了一些插入内容
INSERT INTO Employee
( Id, Name, Department_Id )
VALUES ( 1, 'Peter', 1 ),
( 2, 'Alex', 1 ),
( 3, 'Sam', 2 ),
( 4, 'James', 2 ),
( 5, 'Anna', 3 ),
( 6, 'Susan', 3 ),
( 7, 'Abby', 4 ),
( 8, 'Endy', 4 );
INSERT INTO Department
( Id, DepartmentName, Parent_Id )
VALUES ( 1, 'IT', NULL ),
( 2, 'HR', NULL),
( 3, 'SubIT', 1 ),
( 4, 'SubSubIT', 3 );
INSERT INTO Salary
( Id, Date, Amount, Employee_Id )
VALUES ( 1, '2013-01-09 16:03:50.003', 3000, 1 ),
( 2, '2013-01-11 16:03:50.003', 5000, 2 ),
( 3, '2013-01-09 16:03:50.003', 2000, 3 ),
( 4, '2013-01-11 16:03:50.003', 1000, 4 ),
( 5, '2013-01-09 16:03:50.003', 4000, 5 ),
( 6, '2013-01-11 16:03:50.003', 6000, 6 ),
( 7, '2013-01-09 16:03:50.003', 7000, 7 ),
( 8, '2013-01-13 16:03:50.003', 9000, 8 );
预期结果是：
Department | Average_Salary
__________________________________
IT | ( X1 + X2 + X3 ) / 3
HR | ( Y1 ) / 1
SubIT | ( X2 + X3 ) / 2
SubSubIT | ( X3 ) / 1
其中：
- X1 - IT 部门的平均工资
- X2 - 副IT部门平均工资
- X3 - SubSubIT 部门的平均工资
- Y1 - 平均值 人力资源部工资
这是一种方法
with avg_per_dep as (
select
[Month] = eomonth(s.date), d.Id, d.DepartmentName
, avgDep = avg(s.Amount * 1.0)
from
Salary s
join Employee e on s.Employee_Id = e.Id
join Department d on e.Department_Id = d.Id
group by d.Id, d.DepartmentName, eomonth(s.date)
)
, rcte as (
select
i = Id, Id
, list = cast(',' + cast(Id as varchar(10)) + ',' as varchar(max))
, step = 1
from
Department
union all
select
a.i, b.Id, cast(a.list + cast(b.Id as varchar(10)) + ',' as varchar(max))
, step + 1
from
rcte a
join Department b on a.Id = b.Parent_Id
)
select
d.DepartmentName, c.[Month]
, Average_Salary = avg(c.avgDep)
from
(
select
top 1 with ties i, list
from
rcte
order by row_number() over (partition by i order by step desc)
) t
join avg_per_dep c on t.list like '%,' + cast(c.Id as varchar(10)) + ',%'
join Department d on t.i = d.Id
group by t.i, d.DepartmentName, c.[Month]
输出
DepartmentName [Month] Average_Salary
---------------------------------------------
IT 2013-01-31 5666.666666
HR 2013-01-31 1500.000000
SubIT 2013-01-31 6500.000000
SubSubIT 2013-01-31 8000.000000
想法：
- 计算每个部门的平均工资
- 获取包含递归 CTE 的所有子级的部门列表。
- 合并两个 table 并计算孩子的平均值
示例数据
我添加了几行具有更宽树结构的行。
DECLARE @Employee TABLE
(
Id INT NOT NULL ,
Name VARCHAR(200) NOT NULL ,
Department_Id INT NOT NULL ,
PRIMARY KEY ( Id )
);
DECLARE @Department TABLE
(
Id INT NOT NULL ,
DepartmentName VARCHAR(200) NOT NULL ,
Parent_Id INT ,
PRIMARY KEY ( Id )
);
DECLARE @Salary TABLE
(
Id INT NOT NULL ,
Date DATETIME NOT NULL ,
Amount INT NOT NULL ,
Employee_Id INT NOT NULL ,
PRIMARY KEY ( Id )
);
INSERT INTO @Employee
( Id, Name, Department_Id )
VALUES
( 1, 'Peter', 1 ),
( 2, 'Alex', 1 ),
( 3, 'Sam', 2 ),
( 4, 'James', 2 ),
( 5, 'Anna', 3 ),
( 6, 'Susan', 3 ),
( 7, 'Abby', 4 ),
( 8, 'Endy', 4 ),
(10, 'e_A', 10),
(11, 'e_AB', 11),
(12, 'e_AC', 12),
(13, 'e_AD', 13),
(14, 'e_ACE', 14),
(15, 'e_ACF', 15),
(16, 'e_ACG', 16);
INSERT INTO @Department
( Id, DepartmentName, Parent_Id )
VALUES
( 1, 'IT', NULL ),
( 2, 'HR', NULL),
( 3, 'SubIT', 1 ),
( 4, 'SubSubIT', 3 ),
(10, 'A', NULL ),
(11, 'AB', 10),
(12, 'AC', 10),
(13, 'AD', 10),
(14, 'ACE', 12),
(15, 'ACF', 12),
(16, 'ACG', 12);
INSERT INTO @Salary
( Id, Date, Amount, Employee_Id )
VALUES
( 1, '2013-01-09 16:03:50.003', 3000, 1 ),
( 2, '2013-01-11 16:03:50.003', 5000, 2 ),
( 3, '2013-01-09 16:03:50.003', 2000, 3 ),
( 4, '2013-01-11 16:03:50.003', 1000, 4 ),
( 5, '2013-01-09 16:03:50.003', 4000, 5 ),
( 6, '2013-01-11 16:03:50.003', 6000, 6 ),
( 7, '2013-01-09 16:03:50.003', 7000, 7 ),
( 8, '2013-01-13 16:03:50.003', 9000, 8 ),
(10, '2013-01-13 16:03:50', 100, 10),
(11, '2013-01-13 16:03:50', 100, 11),
(12, '2013-01-13 16:03:50', 100, 12),
(13, '2013-01-13 16:03:50', 100, 13),
(14, '2013-01-13 16:03:50', 100, 14),
(15, '2013-01-13 16:03:50', 100, 15),
(16, '2013-01-13 16:03:50', 100, 16);
查询
WITH
CTE_Departments
AS
(
SELECT
D.Id
,D.Parent_Id
,D.DepartmentName
,SUM(Amount) AS DepartmentAmount
,COUNT(*) AS DepartmentCount
FROM
@Department AS D
INNER JOIN @Employee AS E ON E.Department_Id = D.Id
INNER JOIN @Salary AS S ON S.Employee_Id = E.Id
GROUP BY
D.Id
,D.Parent_Id
,D.DepartmentName
)
,CTE_Recursive
AS
(
SELECT
CTE_Departments.Id AS OriginalID
,CTE_Departments.DepartmentName AS OriginalName
,CTE_Departments.Id
,CTE_Departments.Parent_Id
,CTE_Departments.DepartmentName
,CTE_Departments.DepartmentAmount
,CTE_Departments.DepartmentCount
,1 AS Lvl
FROM CTE_Departments
UNION ALL
SELECT
CTE_Recursive.OriginalID
,CTE_Recursive.OriginalName
,CTE_Departments.Id
,CTE_Departments.Parent_Id
,CTE_Departments.DepartmentName
,CTE_Departments.DepartmentAmount
,CTE_Departments.DepartmentCount
,CTE_Recursive.Lvl + 1 AS Lvl
FROM
CTE_Departments
INNER JOIN CTE_Recursive ON CTE_Recursive.Id = CTE_Departments.Parent_Id
)
SELECT
OriginalID
,OriginalName
,SUM(DepartmentAmount) AS SumAmount
,SUM(DepartmentCount) AS SumCount
,SUM(DepartmentAmount) / SUM(DepartmentCount) AS AvgAmount
FROM CTE_Recursive
GROUP BY
OriginalID
,OriginalName
ORDER BY OriginalID
;
结果
+------------+--------------+-----------+----------+-----------+
| OriginalID | OriginalName | SumAmount | SumCount | AvgAmount |
+------------+--------------+-----------+----------+-----------+
| 1 | IT | 34000 | 6 | 5666 |
| 2 | HR | 3000 | 2 | 1500 |
| 3 | SubIT | 26000 | 4 | 6500 |
| 4 | SubSubIT | 16000 | 2 | 8000 |
| 10 | A | 700 | 7 | 100 |
| 11 | AB | 100 | 1 | 100 |
| 12 | AC | 400 | 4 | 100 |
| 13 | AD | 100 | 1 | 100 |
| 14 | ACE | 100 | 1 | 100 |
| 15 | ACF | 100 | 1 | 100 |
| 16 | ACG | 100 | 1 | 100 |
+------------+--------------+-----------+----------+-----------+
运行 逐步查询，逐个 CTE 以了解其工作原理。
CTE_Departments给出每个部门的总金额和人数。
CTE_Recursive 递归地为每个部门生成子行，同时保持 OriginalID - 递归开始的部门 ID。
最终查询只是按此 OriginalID.
您也可以使用下面的查询来获得预期的结果
WITH Department_Path
AS (SELECT Id, CAST(CONCAT('@', Id, '@') AS VARCHAR(255)) AS Path
FROM Department
WHERE Parent_Id IS NULL
UNION ALL
SELECT Child.Id, CAST(CONCAT(Parent.Path, Child.Id, '@') AS VARCHAR(255)) AS Path
FROM Department Child
INNER JOIN Department_Path Parent
ON Parent.Id = Child.Parent_Id)
SELECT Department.Id,
Department.DepartmentName,
AVG(Salary.Amount) As Average_Salary,
COUNT(Employee.Id) AS Employee_Count
FROM Department
INNER JOIN Department_Path
ON CHARINDEX(CONCAT('@', Department.Id, '@'), Department_Path.Path) > 0
INNER JOIN Employee
ON Employee.Department_Id = Department_Path.Id
INNER JOIN Salary
ON Salary.Employee_Id = Employee.Id
GROUP BY Department.Id,
Department.DepartmentName;
这个想法是每个员工都属于一个层次结构部门列表。对于每个部门，我们可以检索属于它的所有员工，然后计算平均工资。