Above six are the most important concepts related to CTE and SQL Server. There are many more things one has to learn, but without beginners fundamentals one can’t learn the advanced concepts. Let us have small quiz and check how many of you get the fundamentals right. Let us explore that how to build Hierarchical Query using a Recursive CTE

Quiz – Hierarchical Query
1) You have an employee table with the following data.
EmpID | FirstName | LastName | MgrID |
1 | David | Kennson | 11 |
2 | Eric | Bender | 11 |
3 | Lisa | Kendall | 4 |
4 | David | Lonning | 11 |
5 | John | Marshbank | 4 |
6 | James | Newton | 3 |
7 | Sally | Smith | NULL |
You need to write a recursive CTE that shows the EmpID, FirstName, LastName, MgrID, and employee level. The CEO should be listed at Level 1. All people who work for the CEO will be listed at Level 2. All of the people who work for those people will be listed at Level 3. Which CTE code will achieve this result?
- WITH EmpList AS
(SELECT Boss.EmpID, Boss.FName, Boss.LName, Boss.MgrID,
1 AS Lvl
FROM Employee AS Boss WHERE Boss.MgrID IS NULL
UNION ALL
SELECT E.EmpID, E.FirstName, E.LastName, E.MgrID, EmpList.Lvl + 1
FROM Employee AS E INNER JOIN EmpList
ON E.MgrID = EmpList.EmpID)
SELECT * FROM EmpList - WITH EmpListAS
(SELECT EmpID, FirstName, LastName, MgrID, 1 as Lvl
FROM Employee WHERE MgrID IS NULL
UNION ALL
SELECT EmpID, FirstName, LastName, MgrID, 2 as Lvl )
SELECT * FROM BossList - WITH EmpList AS
(SELECT EmpID, FirstName, LastName, MgrID, 1 as Lvl
FROM Employee WHERE MgrID is NOT NULL
UNION
SELECT EmpID, FirstName, LastName, MgrID, BossList.Lvl + 1
FROM Employee INNER JOIN EmpList BossList
ON Employee.MgrID = BossList.EmpID)
SELECT * FROM EmpList
2) You have a table named Employee. The EmployeeID of each employee’s manager is in the ManagerID column. You need to write a recursive query that produces a list of employees and their manager. The query must also include the employee’s level in the hierarchy. You write the following code segment:
WITH EmployeeList (EmployeeID, FullName, ManagerName, Level)
AS (
–PICK ANSWER CODE HERE
)
- SELECT EmployeeID, FullName, ” AS [ManagerID], 1 AS [Level]
FROM Employee
WHERE ManagerID IS NULL
UNION ALL
SELECT emp.EmployeeID, emp.FullName mgr.FullName, 1 + 1 AS [Level] FROM Employee emp JOIN Employee mgr
ON emp.ManagerID = mgr.EmployeeId - SELECT EmployeeID, FullName, ” AS [ManagerID], 1 AS [Level]
FROM Employee
WHERE ManagerID IS NULL
UNION ALL
SELECT emp.EmployeeID, emp.FullName, mgr.FullName, mgr.Level + 1
FROM EmployeeList mgr JOIN Employee emp
ON emp.ManagerID = mgr.EmployeeId
Now make sure that you write down all the answers on the piece of paper.
Watch following video and read the earlier article over here. If you want to change the answer you still have a chance.
Solution
1) 1
2) 2
Now compare let us check the answers and compare your answers to following answers. I am very confident you will get them correct.
Please leave your feedback in the comment area for the quiz and video. Did you know all the answers of the quiz?
Reference: Pinal Dave (https://blog.sqlauthority.com)
5 Comments. Leave new
love it… very informative.
Thanks
can u please any one post the query for bottom to top hierarchy ..
Hi pinal,
I have tried the above example but its not working for me. i tried the first question. but its showing only one record. Here is my defination.
create table #Employee
(
EmpID int identity(1,1),
FirstName varchar(500),
MgrID int
)
insert into #Employee
(
FirstName,
MgrID
)
values
(
‘David’,
11
),(‘Eric’,11),(‘Lisa’,4),(‘David’,11),(‘John’,4),(‘James’,3),(‘Sally’,null)
select * from #Employee
———–Emp with recurssion
WITH EmpList AS
(
SELECT
Boss.EmpID,
Boss.FirstName,
Boss.MgrID,
1 AS Lvl
FROM
#Employee AS Boss
WHERE
MgrID IS NULL
UNION ALL
SELECT
E.EmpID,
E.FirstName,
E.MgrID,
EmpList.Lvl + 1
FROM
#Employee AS E
INNER JOIN EmpList ON
E.MgrID = EmpList.EmpID
)
SELECT distinct * FROM EmpList order by MgrID asc
Thanks,
Question…why would you join Manager ID to employee ID and not employee id to employee id?
alll broken. None of its seems to work.