Difference between CROSS APPLY and OUTER APPLY?

Question

What’s the difference between CROSS APPLY and OUTER APPLY? How do they work? In what situation can they be used? It would be possible to show some examples?

Answer 1

APPLY is a command analogous to JOIN, only for functions (FUNCTIONS).

According to the Technet:

The APPLY operator allows you to invoke a function with table value for each row returned by an external table expression of a query. The table value function acts as the right input and the outer table expression acts as the left input. The right input is evaluated for each line of the left input and the produced lines are combined in the final output. The list of columns produced by the APPLY operator is the set of columns in the left entry, followed by the list of columns returned by the right entry.

There are two ways to apply: CROSS APPLY and OUTER APPLY. CROSS APPLY only returns rows from the outer table that produce a set of function results with table value. OUTER APPLY returns rows that produce a set of results and rows that do not, with NULL values in the columns produced by the function with table value.

Suppose the tables:

--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);

Suppose the following function to retrieve a table subtree Employees:

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

Using the following command:

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;

It is obtained:

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

Answer 2

To summarize: CROSS APPLY - Similar to INNER JOIN
OUTER APPLY - Similar to LEFT JOIN

You can use the fields of the other tables referenced in Where, Functions, Use Top.