SQL Server/T-SQL Tutorial/Table Join/APPLY — различия между версиями

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Версия 16:46, 26 мая 2010

Using APPLY to Invoke a Table-Valued Function for Each Row

   <source lang="sql">

APPLY is used to invoke a table-valued function for each row of an outer query. A table-valued function returns a result set based on one or more parameters. Using APPLY, the input of these parameters are the columns of the left referencing table. CROSS APPLY works like an INNER JOIN. Unmatched rows between the left table and the table-valued function do"n appear in the result set. OUTER APPLY is like an OUTER JOIN. Non-matched rows are still returned in the result set with NULL values in the function results.

14> 15> CREATE TABLE employee( 16> id INTEGER NOT NULL PRIMARY KEY, 17> first_name VARCHAR(10), 18> last_name VARCHAR(10), 19> salary DECIMAL(10,2), 20> start_Date DATETIME, 21> region VARCHAR(10), 22> city VARCHAR(20), 23> managerid INTEGER 24> ); 25> GO 1> INSERT INTO employee VALUES (1, "Jason" , "Martin", 5890,"2005-03-22","North","Vancouver",3); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (2, "Alison", "Mathews",4789,"2003-07-21","South","Utown",4); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (3, "James" , "Smith", 6678,"2001-12-01","North","Paris",5); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (4, "Celia" , "Rice", 5567,"2006-03-03","South","London",6); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (5, "Robert", "Black", 4467,"2004-07-02","East","Newton",7); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (6, "Linda" , "Green" , 6456,"2002-05-19","East","Calgary",8); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (7, "David" , "Larry", 5345,"2008-03-18","West","New York",9); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (8, "James" , "Cat", 4234,"2007-07-17","West","Regina",9); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (9, "Joan" , "Act", 6123,"2001-04-16","North","Toronto",10); 2> GO (1 rows affected) 1> 2> select * from employee; 3> GO id first_name last_name salary start_Date region city managerid


---------- ---------- ------------ ----------------------- ---------- -------------------- -----------
         1 Jason      Martin          5890.00 2005-03-22 00:00:00.000 North      Vancouver                      3
         2 Alison     Mathews         4789.00 2003-07-21 00:00:00.000 South      Utown                          4
         3 James      Smith           6678.00 2001-12-01 00:00:00.000 North      Paris                          5
         4 Celia      Rice            5567.00 2006-03-03 00:00:00.000 South      London                         6
         5 Robert     Black           4467.00 2004-07-02 00:00:00.000 East       Newton                         7
         6 Linda      Green           6456.00 2002-05-19 00:00:00.000 East       Calgary                        8
         7 David      Larry           5345.00 2008-03-18 00:00:00.000 West       New York                       9
         8 James      Cat             4234.00 2007-07-17 00:00:00.000 West       Regina                         9
         9 Joan       Act             6123.00 2001-04-16 00:00:00.000 North      Toronto                       10

(9 rows affected) 1> 2> 3> 4> CREATE FUNCTION dbo.fn_getByID(@ID int) RETURNS TABLE 5> AS 6> RETURN 7> SELECT first_name 8> FROM employee 9> WHERE ID = @ID 10> GO 1> 2> SELECT w.last_name 3> FROM employee w 4> CROSS APPLY dbo.fn_getByID 5> (w.ID) AS r 6> ORDER BY w.ID; 7> GO last_name


Martin Mathews Smith Rice Black Green Larry Cat Act (9 rows affected) 1> 2> drop function dbo.fn_getByID; 3> GO 1> 2> 3> 4> drop table employee; 5> GO</source>