APPLY Operator in T-SQL

APPLY operator was introduced with SQL Server 2005 edition and was great help to join records set with table value function or table value expression.

There are two APPLY operators.

1. CROSS APPLY

2. OUTER APPLY

In simplest terms, CROSS APPLY behave like INNER JOIN and OUTER APPLY behave like LEFT OUTER JOIN.

Let's consider a following example table structure

-- Employee Table

CREATE TABLE Employee (
    EmpId INT PRIMARY KEY,
    EmployeeName VARCHAR(100),
    Age INT,
    BranchId INT
);

-- Branch Table

CREATE TABLE Branch (
    BranchId INT PRIMARY KEY,
    BranchName VARCHAR(100)
);

Here is my test data:

Branch Table

 

Employee Table

I have a table value function which brings branch id and average age of employees in that branch.

CREATE FUNCTION GetAverageEmployeeAgeByBranch
(
    @BranchId INT
)
RETURNS TABLE
AS
RETURN
(
    SELECT BranchId, AVG(Age) AS AverageAge
    FROM Employee
    WHERE BranchId = @BranchId
    GROUP BY BranchId
);

I need to show branch id, branch name and average employee age for that branch in my query. First I have used CROSS apply in my query.

SELECT b.BranchId, BranchName, e.AverageAge

From Branch b CROSS APPLY dbo.GetAverageEmployeeAgeByBranch(BranchId) as e

ORDER BY BranchId

Result as follows:

Then I have used OUTER APPLY:

SELECT b.BranchId, BranchName, e.AverageAge

From Branch b OUTER APPLY dbo.GetAverageEmployeeAgeByBranch(BranchId) as e

ORDER BY BranchId

Result as follows:

As you can see in the first result set, SQL server only brings branches where there are employees on them (i.e. inner join), but on second result set it has bring all branches, where it cannot find employee age, it has put null on them (like LEFT OUTER join).

You can also use APPLY operator with table value expressions (i.e. sub queries).

SELECT b.BranchId, b.BranchName, e.EmployeeName

FROM Branch b CROSS APPLY 

(SELECT e.BranchId, 

STRING_AGG(EmployeeName, ';') WITHIN GROUP (ORDER BY e.EmployeeName) As EmployeeName 

FROM Employee e WHERE e.BranchId = b.BranchId GROUP BY e.BranchId) as e

Above query brings concatenated employee names for a given branch: