Cross apply vs Outer apply

The APPLY operator allows you to join two table expressions; the right table expression is processed every time for each row from the left table expression. As you might have guessed, the left table expression is evaluated first and then right table expression is evaluated against each row of the left table expression for final result-set. The final result-set contains all the selected columns from the left table expression followed by all the columns of right table expression.

The APPLY operator comes in two variants, the CROSS APPLY and the OUTER APPLY. The CROSS APPLY operator returns only those rows from left table expression (in its final output) if it matches with right table expression. In other words, the right table expression returns rows for left table expression match only.  Whereas the OUTER APPLY operator returns all the rows from left table expression irrespective of its match with the right table expression.  For those rows for which there are no corresponding matches in right table expression, it contains NULL values in columns of right table expression. So you might now conclude, the CROSS APPLY is semantically equivalent to INNER JOIN (or to be more precise its like a CROSS JOIN with a correlated sub-query) with a implicit join condition of 1=1 whereas OUTER APPLY is semantically equivalent to LEFT OUTER JOIN.

You might be wondering if the same can be achieved with regular JOIN clause then why and when to use APPLY operator? Though the same can be achieved with normal JOIN, the need of APPLY arises if you have table-valued expression on right part and also in some cases use of APPLY operator boost the performance of your query. Let me explain you with help of some examples.

Script #1 creates a Department table to hold information about departments. Then it creates an Employee table which hold information about the employees. Please note, each employee belongs to a department, hence the Employee table has referential integrity with the Department table.

Script #1 - Creating some temporary objects to work on...

USE [tempdb]  GO IF EXISTS (SELECT * FROM sys.objects WHERE OBJECT_ID = OBJECT_ID(N'[Employee]') AND type IN (N'U')) BEGIN     DROP TABLE [Employee] END GO IF EXISTS (SELECT * FROM sys.objects WHERE OBJECT_ID = OBJECT_ID(N'[Department]') AND type IN (N'U'))BEGIN     DROP TABLE [Department] END  CREATE TABLE [Department](     [DepartmentID] [int] NOT NULL PRIMARY KEY,     [Name] VARCHAR(250) NOT NULL,  ) ON [PRIMARY] INSERT [Department] ([DepartmentID], [Name])  VALUES (1, N'Engineering') INSERT [Department] ([DepartmentID], [Name])  VALUES (2, N'Administration') INSERT [Department] ([DepartmentID], [Name])  VALUES (3, N'Sales') INSERT [Department] ([DepartmentID], [Name])  VALUES (4, N'Marketing') INSERT [Department] ([DepartmentID], [Name])  VALUES (5, N'Finance') GO CREATE TABLE [Employee](     [EmployeeID] [int] NOT NULL PRIMARY KEY,     [FirstName] VARCHAR(250) NOT NULL,     [LastName] VARCHAR(250) NOT NULL,     [DepartmentID] [int] NOT NULL REFERENCES [Department](DepartmentID),  ) ON [PRIMARY]  GO INSERT [Employee] ([EmployeeID], [FirstName], [LastName], [DepartmentID]) VALUES (1, N'Orlando', N'Gee', 1 ) INSERT [Employee] ([EmployeeID], [FirstName], [LastName], [DepartmentID]) VALUES (2, N'Keith', N'Harris', 2 ) INSERT [Employee] ([EmployeeID], [FirstName], [LastName], [DepartmentID]) VALUES (3, N'Donna', N'Carreras', 3 ) INSERT [Employee] ([EmployeeID], [FirstName], [LastName], [DepartmentID]) VALUES (4, N'Janet', N'Gates', 3 )

First query in Script #2 selects data from Department table and uses CROSS APPLY to evaluate the Employee table for each record of the Department table. Second query simply joins the Department table with the Employee table and all the matching records are produced.

Script #2 - CROSS APPLY and INNER JOIN

SELECT * FROM Department D CROSS APPLY     (     SELECT * FROM Employee E     WHERE E.DepartmentID = D.DepartmentID     ) A  GO SELECT * FROM Department D INNER JOIN Employee E ON D.DepartmentID = E.DepartmentID  GO

If you look at the results they produced, it is the exact same result-set; not only that even the execution plan for these queries are similar to each other and has equal query cost, as you can see in the image below. So what is the use of APPLY operator?  How does it differ from a JOIN and how does it help in writing more efficient queries. I will discuss this later, but first let me show you an example of OUTER APPLY also.

The first query in Script #3 selects data from Department table and uses OUTER APPLY to evaluate the Employee table for each record of the Department table.  For those rows for which there is not a match in Employee table, those rows contains NULL values as you can see in case of row 5 and 6. The second query simply uses a LEFT OUTER JOIN between the Department table and the Employee table.  As expected the query returns all rows from Department table; even for those rows for which there is no match in the Employee table.

Script #3 - OUTER APPLY and LEFT OUTER JOIN

SELECT * FROM Department D OUTER APPLY     (     SELECT * FROM Employee E     WHERE E.DepartmentID = D.DepartmentID     ) A  GO SELECT * FROM Department D LEFT OUTER JOIN Employee E ON D.DepartmentID = E.DepartmentID  GO

Even though the above two queries return the same information, the execution plan is a bit different. Although cost wise there is not much difference, the query with the OUTER APPLY uses a Compute Scalar operator (which has an estimated operator cost of 0.0000103 or almost 0% of total query cost) before Nested Loops operator to evaluate and produce the columns of Employee table.

Now comes the time to see where the APPLY operator is really required. In Script #4, I am creating a table-valued function which accepts DepartmentID as its parameter and returns all the employees who belong to this department. The next query selects data from Department table and uses CROSS APPLY to join with the function we created.  It passes the DepartmentID for each row from the outer table expression (in our case Department table) and evaluates the function for each row similar to a correlated subquery. The next query uses the OUTER APPLY in place of CROSS APPLY and hence unlike CROSS APPLY which returned only correlated data, the OUTER APPLY returns non-correlated data as well, placing NULLs into the missing columns.

Script #4 - APPLY with table-valued function

IF EXISTS (SELECT * FROM sys.objects WHERE OBJECT_ID = OBJECT_ID(N'[fn_GetAllEmployeeOfADepartment]') AND type IN (N'IF'))BEGIN     DROP FUNCTION dbo.fn_GetAllEmployeeOfADepartment END GO CREATE FUNCTION dbo.fn_GetAllEmployeeOfADepartment(@DeptID AS INT)  RETURNS TABLE  AS  RETURN     (     SELECT * FROM Employee E     WHERE E.DepartmentID = @DeptID     ) GO SELECT * FROM Department D CROSS APPLY dbo.fn_GetAllEmployeeOfADepartment(D.DepartmentID) GO SELECT * FROM Department D OUTER APPLY dbo.fn_GetAllEmployeeOfADepartment(D.DepartmentID) GO

So now if you are wondering, can we use a simple join in place of the above queries? Then the answer is NO, if you replace CROSS/OUTER APPLY in the above queries with INNER JOIN/LEFT OUTER JOIN, specify ON clause (something as 1=1) and run the query, you will get "The multi-part identifier "D.DepartmentID" could not be bound."  error. This is because with JOINs the execution context of outer query is different from the execution context of the function (or a derived table), and you can not bind a value/variable from the outer query to the function as a parameter.  Hence the APPLY operator is required for such queries.

So in summary the APPLY operator is required when you have to use table-valued function in the query, but it can also be used with an inline SELECT statements.