Problem
SQL Server 2005 introduced the APPLY operator, which is very much like a join clause and which allows joining between two table expressions i.e. joining a left/outer table expression with a right/inner table expression. The difference between join and APPLY operator becomes evident when you have a table-valued expression on the right side and you want this table-valued expression to be evaluated for each row from the left table expression. In this tip I am going to demonstrate what APPLY operator is, how it differs from regular JOINs and what are few of its applications.
Solution
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...
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USE [tempdb] |
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
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SELECT * FROM Department D |
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
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SELECT * FROM Department D |
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
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IF EXISTS (SELECT * FROM sys.objects WHERE OBJECT_ID = OBJECT_ID(N'[fn_GetAllEmployeeOfADepartment]') AND type IN (N'IF'))BEGIN |
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 aderived 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.
Now let me show you another query with a Dynamic Management Function (DMF). Script #5 returns all the currently executing user queries except for the queries being executed by the current session. As you can see the script below, the sys.dm_exec_requests dynamic management view is being CROSS APPLY'ed with the sys.dm_exec_sql_text dynamic management function which accepts a "plan handle" for the query and the same "plan handle" is being passed from the left/outer expression to the function to work and to return the data.
Script #5 - APPLY with Dynamic Management Function (DMF)
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USE master |
Please note the [text] column in the above query returns the all queries submitted in a batch; if you want to see only active (currently executing) query you can use statement_start_offset and statement_end_offset columns to trim the active part of the query. Tim Ford has provided a very good explanation of usage of these columns in his How to isolate the current running commands in SQL Server tip.
As I told you before there are certain scenarios where a query with APPLY operator performs better than a query with regular joins but I am not going to delve into much details rather here are some articles which discuss this topic in greater details.
Please note, APPLY operator is not an ANSI operator but rather an extension of SQL Server T-SQL (available in SQL Server 2005 and above), so if you plan to port your database to some other DBMS take this into your considerations.
Next Steps
- Review Dynamic Management Views\Functions tips.
- Review Splitting Delimited Strings Using XML in SQL Server tip.
- Review SQL Server Monitoring Scripts with the DMVs tip.
- Review Collecting Query Statistics for SQL Server 2005 tip.
- Review my all previous tips.
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.
 Note |
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To use APPLY, the database compatibility level must be at least 90.
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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:
--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.
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.
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.
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.
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