我有以下表格：
Employees
-------------
ClockNo int
CostCentre varchar
Department int
和
Departments
-------------
DepartmentCode int
CostCentreCode varchar
Parent int
部门可以将其他部门作为父母，这意味着存在无限的层次结构。所有部门都属于成本中心，因此总是有CostCentreCode。如果parent = 0它是顶级部门
员工必须具有CostCentre值，但Department可能为0，表示他们不在某个部门
我想要尝试生成的是一个查询，它将提供最多四层次的层次结构。像这样：
EmployeesLevels
-----------------
ClockNo
CostCentre
DeptLevel1
DeptLevel2
DeptLevel3
DeptLevel4
我已经设法在其上显示部门结构，但我无法解决如何将其与员工联系而不会创建重复的员工行：
SELECT d1.Description AS lev1, d2.Description as lev2, d3.Description as lev3, d4.Description as lev4
FROM departments AS d1
LEFT JOIN departments AS d2 ON d2.parent = d1.departmentcode
LEFT JOIN departments AS d3 ON d3.parent = d2.departmentcode
LEFT JOIN departments AS d4 ON d4.parent = d3.departmentcode
WHERE d1.parent=0;
SQL创建Structure和一些示例数据：
CREATE TABLE Employees(
ClockNo integer NOT NULL PRIMARY KEY,
CostCentre varchar(20) NOT NULL,
Department integer NOT NULL);
CREATE TABLE Departments(
DepartmentCode integer NOT NULL PRIMARY KEY,
CostCentreCode varchar(20) NOT NULL,
Parent integer NOT NULL
);
CREATE INDEX idx0 ON Employees (ClockNo);
CREATE INDEX idx1 ON Employees (CostCentre, ClockNo);
CREATE INDEX idx2 ON Employees (CostCentre);
CREATE INDEX idx0 ON Departments (DepartmentCode);
CREATE INDEX idx1 ON Departments (CostCentreCode, DepartmentCode);
INSERT INTO Employees VALUES (1, 'AAA', 0);
INSERT INTO Employees VALUES (2, 'AAA', 3);
INSERT INTO Employees VALUES (3, 'BBB', 0);
INSERT INTO Employees VALUES (4, 'BBB', 4);
INSERT INTO Employees VALUES (5, 'CCC', 0);
INSERT INTO Employees VALUES (6, 'AAA', 1);
INSERT INTO Employees VALUES (7, 'AAA', 5);
INSERT INTO Employees VALUES (8, 'AAA', 15);
INSERT INTO Departments VALUES (1, 'AAA', 0);
INSERT INTO Departments VALUES (2, 'AAA', 1);
INSERT INTO Departments VALUES (3, 'AAA', 1);
INSERT INTO Departments VALUES (4, 'BBB', 0);
INSERT INTO Departments VALUES (5, 'AAA', 3);
INSERT INTO Departments VALUES (12, 'AAA', 5);
INSERT INTO Departments VALUES (15, 'AAA', 12);
这给出了以下结构（方括号中的员工时钟数字）：
Root
|
|---AAA [1]
| \---1 [6]
| |---2
| \---3 [2]
| \---5 [7]
| \---12
| \---15 [8]
|
|---BBB [3]
| \---4 [4]
|
\---CCC [5]
查询应返回以下内容：
ClockNo CostCentre Level1 Level2 Level3 Level4
1 AAA
2 AAA 1 3
3 BBB
4 BBB 4
5 CCC
6 AAA 1
7 AAA 1 3 5
8 AAA 1 3 5 12 *
*对于员工8，他们处于第5级。理想情况下，我希望将所有级别显示为level4，但我很高兴在这种情况下显示CostCentre
答案 0 :(得分：4)
当我们加入表格时，当我们找到属于上一级员工的适当部门时，我们应该停止进一步遍历路径。
当Employee.Department = 0时，我们也有例外情况。在这种情况下，我们不应该加入任何部门，因为在这种情况下，部门是根。
我们只需要在其中一个级别选择包含员工部门的记录。 如果员工的部门级别大于4，我们应该扩展所有4个级别的部门并按原样显示（即使无法达到所需的部门级别并且未在扩展中找到它那些）。
select e.ClockNo,
e.CostCentre,
d1.DepartmentCode as Level1,
d2.DepartmentCode as Level2,
d3.DepartmentCode as Level3,
d4.DepartmentCode as Level4
from Employees e
left join Departments d1
on e.CostCentre=d1.CostCentreCode
and d1.Parent=0
and ((d1.DepartmentCode = 0 and e.Department = 0) or e.Department <> 0)
left join Departments d2
on d2.parent=d1.DepartmentCode
and (d1.DepartMentCode != e.Department and e.Department<>0)
left join Departments d3
on d3.parent=d2.DepartmentCode
and (d2.DepartMentCode != e.Department and e.Department<>0)
left join Departments d4
on d4.parent=d3.DepartmentCode
and (d3.DepartMentCode != e.Department and e.Department<>0)
where e.Department=d1.DepartmentCode
or e.Department=d2.DepartmentCode
or e.Department=d3.DepartmentCode
or e.Department=d4.DepartmentCode
or e.Department=0
or (
(d1.DepartmentCode is not null) and
(d2.DepartmentCode is not null) and
(d3.DepartmentCode is not null) and
(d4.DepartmentCode is not null)
)
order by e.ClockNo;
答案 1 :(得分：2)
此处的主要挑战是员工的部门可能需要显示在 Level1 ， Level2 ， Level3 列中，或 Level4 ，具体取决于层次结构中该部门的上层数。
我建议首先在内部查询中查询每个员工的部门级别数，然后使用该信息将部门代码放在右栏中：
SELECT ClockNo, CostCentre,
CASE LevelCount
WHEN 1 THEN Dep1
WHEN 2 THEN Dep2
WHEN 3 THEN Dep3
ELSE Dep4
END Level1,
CASE LevelCount
WHEN 2 THEN Dep1
WHEN 3 THEN Dep2
WHEN 4 THEN Dep3
END Level2,
CASE LevelCount
WHEN 3 THEN Dep1
WHEN 4 THEN Dep2
END Level3,
CASE LevelCount
WHEN 4 THEN Dep1
END Level4
FROM (SELECT e.ClockNo, e.CostCentre,
CASE WHEN d2.DepartmentCode IS NULL THEN 1
ELSE CASE WHEN d3.DepartmentCode IS NULL THEN 2
ELSE CASE WHEN d4.DepartmentCode IS NULL THEN 3
ELSE 4
END
END
END AS LevelCount,
d1.DepartmentCode Dep1, d2.DepartmentCode Dep2,
d3.DepartmentCode Dep3, d4.DepartmentCode Dep4
FROM Employees e
LEFT JOIN departments AS d1 ON d1.DepartmentCode = e.Department
LEFT JOIN departments AS d2 ON d2.DepartmentCode = d1.Parent
LEFT JOIN departments AS d3 ON d3.DepartmentCode = d2.Parent
LEFT JOIN departments AS d4 ON d4.DepartmentCode = d3.Parent) AS Base
ORDER BY ClockNo
或者，您可以根据现有级别（0到4个部门的链）对5种可能的场景进行简单UNION ALL：
SELECT ClockNo, CostCentre, d4.DepartmentCode Level1,
d3.DepartmentCode Level2, d2.DepartmentCode Level3,
d1.DepartmentCode Level4
FROM Employees e
INNER JOIN departments AS d1 ON d1.DepartmentCode = e.Department
INNER JOIN departments AS d2 ON d2.DepartmentCode = d1.Parent
INNER JOIN departments AS d3 ON d3.DepartmentCode = d2.Parent
INNER JOIN departments AS d4 ON d4.DepartmentCode = d3.Parent
UNION ALL
SELECT ClockNo, CostCentre, d3.DepartmentCode,
d2.DepartmentCode, d1.DepartmentCode, NULL
FROM Employees e
INNER JOIN departments AS d1 ON d1.DepartmentCode = e.Department
INNER JOIN departments AS d2 ON d2.DepartmentCode = d1.Parent
INNER JOIN departments AS d3 ON d3.DepartmentCode = d2.Parent
WHERE d3.Parent = 0
UNION ALL
SELECT ClockNo, CostCentre, d2.DepartmentCode,
d1.DepartmentCode, NULL, NULL
FROM Employees e
INNER JOIN departments AS d1 ON d1.DepartmentCode = e.Department
INNER JOIN departments AS d2 ON d2.DepartmentCode = d1.Parent
WHERE d2.Parent = 0
UNION ALL
SELECT ClockNo, CostCentre, d1.DepartmentCode Level1,
NULL, NULL, NULL
FROM Employees e
INNER JOIN departments AS d1 ON d1.DepartmentCode = e.Department
WHERE d1.Parent = 0
UNION ALL
SELECT ClockNo, CostCentre, NULL, NULL, NULL, NULL
FROM Employees e
WHERE e.Department = 0
ORDER BY ClockNo
答案 2 :(得分：2)
SELECT [ClockNo]
, [CostCentre]
, CASE
WHEN Department <> 0 THEN dept.[Level1]
END AS [Level1]
, CASE
WHEN Department <> 0 THEN dept.[Level2]
END AS [Level2]
, CASE
WHEN Department <> 0 THEN dept.[Level3]
END AS [Level3]
, CASE
WHEN Department <> 0 THEN dept.[Level4]
END AS [Level4]
FROM [Employees] emp
LEFT JOIN
(
SELECT
CASE
WHEN d4.[DepartmentCode] IS NOT NULL THEN d4.[DepartmentCode]
WHEN d3.[DepartmentCode] IS NOT NULL THEN d3.[DepartmentCode]
WHEN d2.[DepartmentCode] IS NOT NULL THEN d2.[DepartmentCode]
ELSE d1.[DepartmentCode]
END AS [Level1]
, CASE
WHEN d4.[DepartmentCode] IS NOT NULL THEN d3.[DepartmentCode]
WHEN d3.[DepartmentCode] IS NOT NULL THEN d2.[DepartmentCode]
WHEN d2.[DepartmentCode] IS NOT NULL THEN d1.[DepartmentCode]
ELSE NULL
END AS [Level2]
, CASE
WHEN d4.[DepartmentCode] IS NOT NULL THEN d2.[DepartmentCode]
WHEN d3.[DepartmentCode] IS NOT NULL THEN d1.[DepartmentCode]
ELSE NULL
END AS [Level3]
, CASE
WHEN d4.[DepartmentCode] IS NOT NULL THEN d1.[DepartmentCode]
ELSE NULL
END AS [Level4]
, d1.[DepartmentCode] AS [DepartmentCode]
, d1.[CostCentreCode] AS [CostCenter]
FROM [Departments] d1
LEFT JOIN
[Departments] d2
ON d1.[Parent] = d2.[DepartmentCode]
LEFT JOIN
[Departments] d3
ON d2.[Parent] = d3.[DepartmentCode]
LEFT JOIN
[Departments] d4
ON d3.[Parent] = d4.[DepartmentCode]
) AS dept
ON emp.[Department] = dept.[DepartmentCode]
ORDER BY emp.[ClockNo]
答案 3 :(得分：1)
尝试此查询。不确定如何使用COALESCE来显示大型数据的性能。
我们的想法是构建一个派生的层次结构表，通向每个Department
lev1 lev2 lev3 lev4
1 NULL NULL NULL
1 2 NULL NULL
1 3 NULL NULL
1 3 5 NULL
4 NULL NULL NULL
然后使用最右边的部门与Employees一起加入。这是完整的查询：
SELECT
ClockNo,
CostCentre,
lev1,
lev2,
lev3,
lev4
FROM Employees
LEFT JOIN
(
SELECT
d1.DepartmentCode AS lev1,
NULL as lev2,
NULL as lev3,
NULL as lev4
FROM departments AS d1
WHERE d1.parent=0
UNION ALL
SELECT
d1.DepartmentCode AS lev1,
d2.DepartmentCode as lev2,
NULL as lev3,
NULL as lev4
FROM departments AS d1
JOIN departments AS d2 ON d2.parent = d1.departmentcode
WHERE d1.parent=0
UNION ALL
SELECT
d1.DepartmentCode AS lev1,
d2.DepartmentCode as lev2,
d3.DepartmentCode as lev3,
NULL as lev4
FROM departments AS d1
JOIN departments AS d2 ON d2.parent = d1.departmentcode
JOIN departments AS d3 ON d3.parent = d2.departmentcode
WHERE d1.parent=0
UNION ALL
SELECT
d1.DepartmentCode AS lev1,
d2.DepartmentCode as lev2,
d3.DepartmentCode as lev3,
d4.DepartmentCode as lev4
FROM departments AS d1
JOIN departments AS d2 ON d2.parent = d1.departmentcode
JOIN departments AS d3 ON d3.parent = d2.departmentcode
JOIN departments AS d4 ON d4.parent = d3.departmentcode
WHERE d1.parent=0
) Department
ON COALESCE(Department.lev4, Department.lev3, Department.lev2, Department.lev1) = Employees.Department
ORDER BY ClockNo
答案 4 :(得分：1)
SunnyMagadan的查询很好。但是，根据部门中的员工数量，您可能希望尝试以下方法，使DB优化器有机会仅为部门遍历部门层次结构而不是为部门中的每个员工重复一次。
SELECT e.ClockNo, e.CostCentre, Level1, Level2, Level3, Level4
FROM Employees e
LEFT JOIN
(SELECT
d1.departmentcode
, d1.CostCentreCode
, coalesce (d4.departmentcode, d3.departmentcode
, d2.departmentcode, d1.departmentcode) AS Level1
, case when d4.departmentcode is not null then d3.departmentcode
when d3.departmentcode is not null then d2.departmentcode
when d2.departmentcode is not null then d1.departmentcode end as Level2
, case when d4.departmentcode is not null then d2.departmentcode
when d3.departmentcode is not null then d1.departmentcode end as Level3
, case when d4.departmentcode is not null then d1.departmentcode end as Level4
FROM departments AS d1
LEFT JOIN departments AS d2 ON d1.parent = d2.departmentcode
LEFT JOIN departments AS d3 ON d2.parent = d3.departmentcode
LEFT JOIN departments AS d4 ON d3.parent = d4.departmentcode) d
ON d.DepartmentCode = e.Department AND d.CostCentreCode = e.CostCentre
;
编辑关于5级以上的部门。
任何固定步骤查询都无法获得前4个级别。因此，更改上面的查询只是为了标记它们，例如-1。
, case when d4.Parent > 0 then NULL else
coalesce (d4.departmentcode, d3.departmentcode
, d2.departmentcode, d1.departmentcode) end AS Level1
等等。
答案 5 :(得分：0)
我建议你分开查询以获取员工并获得他/她的部门层级。
为了获得部门的层次结构，我建议你使用这样的递归CTE：
with DepartmentList (DepartmentCode, CostCentreCode, Parent) AS
(
SELECT
parentDepartment.DepartmentCode,
parentDepartment.CostCentreCode,
parentDepartment.Parent
FROM Departments parentDepartment
WHERE DepartmentCode = @departmentCode
UNION ALL
SELECT
childDepartment.DepartmentCode
childDepartment.CostCentreCode,
childDepartment.Parent,
FROM Departments childDepartment
JOIN DepartmentList
ON childDepartment.Parent = DepartmentList.DepartmentCode
)
SELECT * FROM DepartmentList
这不是你问题的直接答案，但这会给你选择和想法。希望这可以帮助。
答案 6 :(得分：0)
所以我已经采取了两个步骤来完成这项工作：
此递归查询构建DepartmentLevels：
;WITH CTE (DepartmentCode, CostCentreCode, Parent, DepartmentLevel)
AS (
SELECT D.DepartmentCode, D.CostCentreCode, D.Parent, 1
FROM dbo.Departments AS D
WHERE D.Parent = 0
UNION ALL
SELECT D.DepartmentCode, D.CostCentreCode, D.Parent, C.DepartmentLevel + 1
FROM dbo.Departments AS D
INNER JOIN CTE AS C
ON C.DepartmentCode = D.Parent
AND C.CostCentreCode = D.CostCentreCode
)
SELECT *
INTO #DepartmentLevels
FROM CTE;
输出结果：
╔════════════════╦════════════════╦════════╦═════════════════╗
║ DepartmentCode ║ CostCentreCode ║ Parent ║ DepartmentLevel ║
╠════════════════╬════════════════╬════════╬═════════════════╣
║ 1 ║ AAA ║ 0 ║ 1 ║
║ 4 ║ BBB ║ 0 ║ 1 ║
║ 2 ║ AAA ║ 1 ║ 2 ║
║ 3 ║ AAA ║ 1 ║ 2 ║
║ 5 ║ AAA ║ 3 ║ 3 ║
╚════════════════╩════════════════╩════════╩═════════════════╝
现在，此查询将为每个节点生成所有可能的父节点（一种映射表）：
;WITH CTE (DepartmentCode, CostCentreCode, Parent, DepartmentLevelCode)
AS (
SELECT D.DepartmentCode, D.CostCentreCode, D.Parent, D.DepartmentCode
FROM dbo.Departments AS D
UNION ALL
SELECT D.DepartmentCode, D.CostCentreCode, D.Parent, C.DepartmentLevelCode
FROM dbo.Departments AS D
INNER JOIN CTE AS C
ON C.Parent = D.DepartmentCode
)
SELECT *
FROM CTE;
这给了我们这个结果：
╔════════════════╦════════════════╦════════╦═════════════════════╗
║ DepartmentCode ║ CostCentreCode ║ Parent ║ DepartmentLevelCode ║
╠════════════════╬════════════════╬════════╬═════════════════════╣
║ 1 ║ AAA ║ 0 ║ 1 ║
║ 2 ║ AAA ║ 1 ║ 2 ║
║ 3 ║ AAA ║ 1 ║ 3 ║
║ 4 ║ BBB ║ 0 ║ 4 ║
║ 5 ║ AAA ║ 3 ║ 5 ║
║ 3 ║ AAA ║ 1 ║ 5 ║
║ 1 ║ AAA ║ 0 ║ 5 ║
║ 1 ║ AAA ║ 0 ║ 3 ║
║ 1 ║ AAA ║ 0 ║ 2 ║
╚════════════════╩════════════════╩════════╩═════════════════════╝
现在我们可以将这三个伙伴与Employees表结合起来，得到所需的输出：
;WITH CTE (DepartmentCode, CostCentreCode, Parent, DepartmentLevelCode)
AS (
SELECT D.DepartmentCode, D.CostCentreCode, D.Parent, D.DepartmentCode
FROM dbo.Departments AS D
UNION ALL
SELECT D.DepartmentCode, D.CostCentreCode, D.Parent, C.DepartmentLevelCode
FROM dbo.Departments AS D
INNER JOIN CTE AS C
ON C.Parent = D.DepartmentCode
)
SELECT E.ClockNo
, E.CostCentre
, C.Level1
, C.Level2
, C.Level3
, C.Level4
FROM dbo.Employees AS E
OUTER APPLY (
SELECT MAX(CASE WHEN DL.DepartmentLevel = 1 THEN C.DepartmentCode END)
, MAX(CASE WHEN DL.DepartmentLevel = 2 THEN C.DepartmentCode END)
, MAX(CASE WHEN DL.DepartmentLevel = 3 THEN C.DepartmentCode END)
, MAX(CASE WHEN DL.DepartmentLevel = 4 THEN C.DepartmentCode END)
FROM CTE AS C
INNER JOIN #DepartmentLevels AS DL
ON DL.DepartmentCode = C.DepartmentCode
WHERE C.DepartmentLevelCode = E.Department
) AS C(Level1, Level2, Level3, Level4);
它会给出这个：
╔═════════╦════════════╦════════╦════════╦════════╦════════╗
║ ClockNo ║ CostCentre ║ Level1 ║ Level2 ║ Level3 ║ Level4 ║
╠═════════╬════════════╬════════╬════════╬════════╬════════╣
║ 1 ║ AAA ║ ║ ║ ║ ║
║ 2 ║ AAA ║ 1 ║ 3 ║ ║ ║
║ 3 ║ BBB ║ ║ ║ ║ ║
║ 4 ║ BBB ║ 4 ║ ║ ║ ║
║ 5 ║ CCC ║ ║ ║ ║ ║
║ 6 ║ AAA ║ 1 ║ ║ ║ ║
║ 7 ║ AAA ║ 1 ║ 3 ║ 5 ║ ║
╚═════════╩════════════╩════════╩════════╩════════╩════════╝
此查询会根据DepartmentLevelCode找到相应的DepartmentCode，并会根据DepartmentLevel来调整内容。希望它是对的。