SQL层次结构和联接表计数

我有一个雇员表和一个项目表。员工只能在1个项目中，并将员工表作为ID和ReportsTO ID的层次结构。我需要得到项目的数量，其中经理的数量将是他手下员工的总和，依此类推

    DECLARE @Employees TABLE (EmpID INT, EmpName VARCHAR(20), ReportsTo INT)
INSERT INTO @Employees(EmpID, EmpName, ReportsTo)
  SELECT 1, 'Jacob', NULL UNION ALL
  SELECT 2, 'Rui', NULL UNION ALL
  SELECT 3, 'Jacobson', NULL UNION ALL
  SELECT 4, 'Jess', 1 UNION ALL
  SELECT 5, 'Steve', 1 UNION ALL
  SELECT 6, 'Bob', 1 UNION ALL
  SELECT 7, 'Smith', 2 UNION ALL
  SELECT 8, 'Bobbey', 2 UNION ALL
  SELECT 9, 'Steffi', 3 UNION ALL
  SELECT 10, 'Bracha', 3 UNION ALL
  SELECT 11, 'John', 5 UNION ALL
  SELECT 12, 'Michael', 6 UNION ALL
  SELECT 13, 'Paul', 6 UNION ALL
  SELECT 14, 'Lana', 7 UNION ALL
  SELECT 15, 'Johnson', 7 UNION ALL
  SELECT 16, 'Mic', 8 UNION ALL
  SELECT 17, 'Stev', 8 UNION ALL
  SELECT 18, 'Paulson', 9 UNION ALL
  SELECT 19, 'Jessica', 10

DECLARE @Projects Table (ID INT,EMPID INT, PROJNAME VARCHAR(50))
INSERT INTO @PROJECTS(ID,EMPID,PROJNAME)  
    SELECT 1,19,'ABCD' UNION ALL
    SELECT 2,5,'EFGH'   UNION ALL

    SELECT 3,16,'ADAD'  UNION ALL

    SELECT 4,6,'FSDFSD' UNION ALL
    SELECT 6,8,'DSDSDSD' UNION ALL
    SELECT 6,8,'DSDSADDSD'



--Solution starts here
DECLARE @manager VARCHAR(20)
SELECT @manager ='JACOB'

--CTE
;With Hierarchy(EmpName, EmpID, Level, FullyQualifiedName)
As
(
  Select E.EmpName, E.EmpID, 0, Cast('.'+E.EmpName+'.' as Varchar(MAX))
    From @Employees E
    Where E.ReportsTo is null
  Union all
  Select E.EmpName, E.EmpID, H.Level+1, H.FullyQualifiedName+'.'+E.EmpName+'.'
    from @Employees E
    inner join Hierarchy H on H.EmpID=E.ReportsTo
)

--Result
Select Space((Level-(Select Top(1) Level from Hierarchy H2 Where H2.EmpName=@manager))*4) + EmpName
  from Hierarchy H
  where CHARINDEX('.'+(Select Top(1) E.EmpName from @Employees E Where E.EmpName=@manager)+'.', H.FullyQualifiedName) > 0
  order by H.FullyQualifiedName

---

如何获得每位员工的项目数量

经理对所有员工的项目总数

--Result
Select Space((Level-(Select Top(1) Level from Hierarchy H2 Where H2.EmpName=@manager))*4) + EmpName,
       COALESCE(CASE WHEN ID IS NOT NULL THEN 1 END,
       CASE WHEN Level = 0 THEN SUM(CASE WHEN ID IS NOT NULL THEN 1 END) OVER() END, 0) AS cnt
from Hierarchy H LEFT JOIN @PROJECTS p ON H.EMPID = p.EMPID
where CHARINDEX('.'+(Select Top(1) E.EmpName from @Employees E Where E.EmpName=@manager)+'.', H.FullyQualifiedName) > 0
order by H.FullyQualifiedName

演示

或者试试这个

--Result
Select Space((Level-(Select Top(1) Level from Hierarchy H2 Where H2.EmpName=@manager))*4) + EmpName,       
       CASE WHEN Level = 0 THEN SUM(CASE WHEN ID IS NOT NULL THEN 1 END) OVER() END AS totalCnt,
       CASE WHEN ID IS NOT NULL THEN 1 END AS cntSingle,              
       o.cntChild
from Hierarchy H OUTER APPLY (
                              SELECT COUNT(p.PROJNAME) AS cntChild
                              FROM @Employees e LEFT JOIN @PROJECTS p ON e.EmpID = p.EMPID
                              WHERE h.EmpID = e.ReportsTo
                              ) o
                 LEFT JOIN @PROJECTS p ON H.EMPID = p.EMPID                           
where CHARINDEX('.'+(Select Top(1) E.EmpName from @Employees E Where E.EmpName=@manager)+'.', H.FullyQualifiedName) > 0
order by H.FullyQualifiedName

---

很好，但是第二级ex：bob必须有孩子总数，但是他的孩子没有项目如果你添加一个或两个孩子，它仍然显示1。我们需要按级别对其进行分区吗？声明@Projects Table ID INT，EMPID INT，PROJNAME VARCHAR50插入@PROJECTSID，EMPID，PROJNAME SELECT 1,19，'ABCD'UNION ALL SELECT 2,5，'EFGH'UNION ALL SELECT 3,16，'ADAD'UNION ALL SELECT 4,6，'FSDFSD'UNION ALL SELECT 6,8，'dsd'UNION ALL SELECT 6,8，“DSDSADDSD”UNION ALL SELECT 7,20，“55555”UNION ALL SELECT 55,12，“sdsds”UNION ALL SELECT 22,13，“dfgdfg”无列名cnt Jacob 6 Bob 1 Michael 1 Paul 1 Jess NULL Steve 1 John 1是分层的多层结构吗？未知性质的X级？什么是rdbms？