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2 User-Defined Functions
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empname VARCHAR(25) NOT NULL, salary MONEY NOT NULL, lvl INT NOT NULL, UNIQUE CLUSTERED(lvl, empid) ) AS BEGIN DECLARE @lvl AS INT; SET @lvl = 0;
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-- Init level counter with 0
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-- Insert root node to @Subs INSERT INTO @Subs(empid, mgrid, empname, salary, lvl) SELECT empid, mgrid, empname, salary, @lvl FROM dbo.Employees WHERE empid = @mgrid; WHILE @@rowcount > 0 BEGIN SET @lvl = @lvl + 1; -- While prev level had rows -- Increment level counter
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-- Insert next level of subordinates to @Subs INSERT INTO @Subs(empid, mgrid, empname, salary, lvl) SELECT C.empid, C.mgrid, C.empname, C.salary, @lvl FROM @Subs AS P -- P = Parent JOIN dbo.Employees AS C -- C = Child ON P.lvl = @lvl - 1 -- Filter parents from prev level AND C.mgrid = P.empid; END RETURN; END GO
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The function accepts the @mgrid input parameter, which is the ID of the input manager . The function returns the @Subs table variable, with details about the input manager and all its subordinates in all levels . In addition to the employee attributes, @Subs also has a column called lvl that keeps track of the level distance from the input manager (0 for the input manager and increasing by one unit for each level) . The function keeps track of the current level in the @lvl local variable, which is initialized with zero . The function first inserts into @Subs the row from Employees with ID equal to @mgrid . Then, in a loop, if the last insert affected more than zero rows, the code increments the @lvl variable s value by one and inserts the next level of employees in other words, direct subordinates of the managers found in the previous level into @Subs . The lvl column is important because it allows you to isolate the employees who were inserted into @Subs in the last iteration . To return only subordinates of the employees found in the previous level, the join condition filters from @Subs only rows where the lvl column is equal to the previous level (@lvl 1) .
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To test the function, run the following code, which returns information about employee 3 and her subordinates:
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SELECT empid, mgrid, empname, salary, lvl FROM dbo.Subordinates(3) AS S;
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This generates the following output:
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empid -----3 7 9 11 12 13 14 mgrid -----1 3 7 7 9 9 9 empname -------Ina Aaron Rita Gabriel Emilia Michael Didi salary -------7500.00 5000.00 3000.00 3000.00 2000.00 2000.00 1500.00 lvl ---0 1 2 2 3 3 3
Interestingly, the task of returning subordinates can also be achieved with an inline table-valued UDF that is based on a recursive query, like so:
IF OBJECT_ID('dbo.Subordinates') IS NOT NULL DROP FUNCTION dbo.Subordinates; GO CREATE FUNCTION dbo.Subordinates(@mgrid AS INT) RETURNS TABLE AS RETURN WITH Subs AS ( -- Anchor member returns a row for the input manager SELECT empid, mgrid, empname, salary, 0 AS lvl FROM dbo.Employees WHERE empid = @mgrid UNION ALL -- Recursive member returns next level of children SELECT C.empid, C.mgrid, C.empname, C.salary, P.lvl + 1 FROM Subs AS P JOIN dbo.Employees AS C ON C.mgrid = P.empid ) SELECT * FROM Subs; GO
The inline UDF applies logic similar to the multistatement UDF, except that it uses recursive common table expressions (CTEs) . It s simpler in the sense that you don t need to define the returned table explicitly or filter the previous level s managers . The first query in the CTE s body returns the row from Employees for the given root employee . It also returns zero as the level of the root employee . In a recursive CTE, a query that doesn t have any recursive references is known as an anchor member .
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