division in SQL results in 0

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declare @x int, @y int, @f float
set @x = 100
set @y = 2000
set @f = @x / (@y * 1.0)
print @x
print @y
print @f

SQL : Apply (Cross Apply / Outer Apply)

Reference: http://technet.microsoft.com/en-us/library/ms175156(v=sql.105).aspx

Using APPLY
SQL Server 2008 R2
Other Versions <javascript:;>


* SQL Server 2008<http://technet.microsoft.com/en-us/library/ms175156(v=sql.100).aspx>
* SQL Server 2005<http://technet.microsoft.com/en-us/library/ms175156(v=sql.90).aspx>
17 out of 32 rated this helpful - Rate this topic<http://technet.microsoft.com/en-us/library/ms175156(v=sql.105).aspx#feedback>

The APPLY operator allows you to invoke a table-valued function for each row returned by an outer table expression of a query. The table-valued function acts as the right input and the outer table expression acts as the left input. The right input is evaluated for each row from the left input and the rows produced are combined for the final output. The list of columns produced by the APPLY operator is the set of columns in the left input followed by the list of columns returned by the right input.
[cid:image001.gif@01CD4B0F.8C893F90]Note


To use APPLY, the database compatibility level must be at least 90.


There are two forms of APPLY: CROSS APPLY and OUTER APPLY. CROSS APPLY returns only rows from the outer table that produce a result set from the table-valued function. OUTER APPLY returns both rows that produce a result set, and rows that do not, with NULL values in the columns produced by the table-valued function.

As an example, consider the following tables, Employees and Departments:
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--Create Employees table and insert values.

CREATE TABLE Employees

(

empid int NOT NULL

,mgrid int NULL

,empname varchar(25) NOT NULL

,salary money NOT NULL

CONSTRAINT PK_Employees PRIMARY KEY(empid)

);

GO

INSERT INTO Employees VALUES(1 , NULL, 'Nancy' , $10000.00);

INSERT INTO Employees VALUES(2 , 1 , 'Andrew' , $5000.00);

INSERT INTO Employees VALUES(3 , 1 , 'Janet' , $5000.00);

INSERT INTO Employees VALUES(4 , 1 , 'Margaret', $5000.00);

INSERT INTO Employees VALUES(5 , 2 , 'Steven' , $2500.00);

INSERT INTO Employees VALUES(6 , 2 , 'Michael' , $2500.00);

INSERT INTO Employees VALUES(7 , 3 , 'Robert' , $2500.00);

INSERT INTO Employees VALUES(8 , 3 , 'Laura' , $2500.00);

INSERT INTO Employees VALUES(9 , 3 , 'Ann' , $2500.00);

INSERT INTO Employees VALUES(10, 4 , 'Ina' , $2500.00);

INSERT INTO Employees VALUES(11, 7 , 'David' , $2000.00);

INSERT INTO Employees VALUES(12, 7 , 'Ron' , $2000.00);

INSERT INTO Employees VALUES(13, 7 , 'Dan' , $2000.00);

INSERT INTO Employees VALUES(14, 11 , 'James' , $1500.00);

GO

--Create Departments table and insert values.

CREATE TABLE Departments

(

deptid INT NOT NULL PRIMARY KEY

,deptname VARCHAR(25) NOT NULL

,deptmgrid INT NULL REFERENCES Employees

);

GO

INSERT INTO Departments VALUES(1, 'HR', 2);

INSERT INTO Departments VALUES(2, 'Marketing', 7);

INSERT INTO Departments VALUES(3, 'Finance', 8);

INSERT INTO Departments VALUES(4, 'R&D', 9);

INSERT INTO Departments VALUES(5, 'Training', 4);

INSERT INTO Departments VALUES(6, 'Gardening', NULL);

Most departments in the Departments table have a manager ID that corresponds to an employee in the Employees table. The following table-valued function accepts an employee ID as an argument and returns that employee and all of his or her subordinates.
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CREATE FUNCTION dbo.fn_getsubtree(@empid AS INT)

RETURNS @TREE TABLE

(

empid INT NOT NULL

,empname VARCHAR(25) NOT NULL

,mgrid INT NULL

,lvl INT NOT NULL

)

AS

BEGIN

WITH Employees_Subtree(empid, empname, mgrid, lvl)

AS

(

-- Anchor Member (AM)

SELECT empid, empname, mgrid, 0

FROM Employees

WHERE empid = @empid



UNION all



-- Recursive Member (RM)

SELECT e.empid, e.empname, e.mgrid, es.lvl+1

FROM Employees AS e

JOIN Employees_Subtree AS es

ON e.mgrid = es.empid

)

INSERT INTO @TREE

SELECT * FROM Employees_Subtree;



RETURN

END

GO

To return all of the subordinates in all levels for the manager of each department, use the following query.
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SELECT D.deptid, D.deptname, D.deptmgrid

,ST.empid, ST.empname, ST.mgrid

FROM Departments AS D

CROSS APPLY fn_getsubtree(D.deptmgrid) AS ST;

Here is the result set.
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deptid deptname deptmgrid empid empname mgrid lvl

----------- ---------- ----------- ----------- ---------- ----------- ---

1 HR 2 2 Andrew 1 0

1 HR 2 5 Steven 2 1

1 HR 2 6 Michael 2 1

2 Marketing 7 7 Robert 3 0

2 Marketing 7 11 David 7 1

2 Marketing 7 12 Ron 7 1

2 Marketing 7 13 Dan 7 1

2 Marketing 7 14 James 11 2

3 Finance 8 8 Laura 3 0

4 R&D 9 9 Ann 3 0

5 Training 4 4 Margaret 1 0

5 Training 4 10 Ina 4 1

Notice that each row from the Departments table is duplicated as many times as there are rows returned from fn_getsubtree for the department's manager.

Also, the Gardening department does not appear in the results. Because this department has no manager, fn_getsubtree returned an empty set for it. By using OUTER APPLY, the Gardening department will also appear in the result set, with null values in the deptmgrid field, as well as in the fields returned by fn_getsubtree.