介绍SQL Server 2005的CROSS Apply

Cross Apply使表可以和表值函数结果进行join, 这样表值函数的参数就可以使用一个结果集，而不是一个标量值，下面是book online的原文，有例子，有解释。

 

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.

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 valuesCREATE TABLE Employees(  empid   int         NOT NULL,  mgrid   int         NULL,  empname varchar(25) NOT NULL,  salary  money       NOT NULL,  CONSTRAINT PK_Employees PRIMARY KEY(empid),)GOINSERT 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 valuesCREATE TABLE Departments(  deptid    INT NOT NULL PRIMARY KEY,  deptname  VARCHAR(25) NOT NULL,  deptmgrid INT NULL REFERENCES Employees)GOINSERT 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 who 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)ASBEGIN  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  RETURNENDGO

To return all of the subordinates in all levels for the manager of each department, use the following query:

 

SELECT *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           01           HR         2           5           Steven     2           11           HR         2           6           Michael    2           12           Marketing  7           7           Robert     3           02           Marketing  7           11          David      7           12           Marketing  7           12          Ron        7           12           Marketing  7           13          Dan        7           12           Marketing  7           14          James      11          23           Finance    8           8           Laura      3           04           R&D        9           9           Ann        3           05           Training   4           4           Margaret   1           05           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.