SQL Server/T-SQL Tutorial/Transact SQL/Variable Select

Материал из SQL эксперт
Перейти к: навигация, поиск

An alternate syntax for setting a variable"s value in a select list

   <source lang="sql">

SELECT @variable_name = column_specification</source>


Assignment with a Subquery

   <source lang="sql">

4> 5> 6> CREATE TABLE employee( 7> id INTEGER NOT NULL PRIMARY KEY, 8> first_name VARCHAR(10), 9> last_name VARCHAR(10), 10> salary DECIMAL(10,2), 11> start_Date DATETIME, 12> region VARCHAR(10), 13> city VARCHAR(20), 14> managerid INTEGER 15> ); 16> 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> DECLARE 3> @ID int 4> SELECT 5> @ID = 0 6> SELECT 7> @ID = 8> ( 9> SELECT 10> ID 11> FROM 12> Employee 13> WHERE 14> ID = 1 15> ) 16> SELECT 17> @ID 18> 19> 20> 21> drop table employee; 22> GO


         1

(1 rows affected) 1></source>


Creating a Comma Delimited List using SELECT

   <source lang="sql">

5> CREATE TABLE employee( 6> id INTEGER NOT NULL PRIMARY KEY, 7> first_name VARCHAR(10), 8> last_name VARCHAR(10), 9> salary DECIMAL(10,2), 10> start_Date DATETIME, 11> region VARCHAR(10), 12> city VARCHAR(20) 13> ); 14> GO 1> INSERT INTO employee VALUES (1, "Jason" , "Martin", 5890,"2005-03-22","North","Vancouver"); 2> GO (1 rows affected) 1> 2> 3> INSERT INTO employee VALUES (2, "Alison", "Mathews",4789,"2003-07-21","South","Utown"); 4> GO (1 rows affected) 1> 2> 3> INSERT INTO employee VALUES (3, "James" , "Smith", 6678,"2001-12-01","North","Paris"); 4> GO (1 rows affected) 1> 2> 3> INSERT INTO employee VALUES (4, "Celia" , "Rice", 5567,"2006-03-03","South","London"); 4> GO (1 rows affected) 1> 2> 3> INSERT INTO employee VALUES (5, "Robert", "Black", 4467,"2004-07-02","East","Newton"); 4> GO (1 rows affected) 1> 2> 3> INSERT INTO employee VALUES (6, "Linda" , "Green" , 6456,"2002-05-19","East","Calgary"); 4> GO (1 rows affected) 1> 2> 3> INSERT INTO employee VALUES (7, "David" , "Larry", 5345,"2008-03-18","West","New York"); 4> GO (1 rows affected) 1> 2> 3> INSERT INTO employee VALUES (8, "James" , "Cat", 4234,"2007-07-17","West","Regina"); 4> GO (1 rows affected) 1> 2> 3> INSERT INTO employee VALUES (9, "Joan" , "Act", 6123,"2001-04-16","North","Toronto"); 4> GO (1 rows affected) 1> 2> 3> 4> select * from employee; 5> GO id first_name last_name salary start_Date region city


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

(9 rows affected) 1> 2> 3> 4> DECLARE @Shifts varchar(20) 5> SET @Shifts = "" 6> SELECT @Shifts = @Shifts + s.first_Name + "," 7> FROM employee s 8> ORDER BY s.start_Date; 9> 10> SELECT @Shifts; 11> GO


Joan,James,Linda,Ali (1 rows affected) 1> 2> 3> 4> drop table employee; 5> GO</source>


Example of a Variable Not Updated Due to No Rows Being Returned

   <source lang="sql">

9> CREATE TABLE employee( 10> id INTEGER NOT NULL PRIMARY KEY, 11> first_name VARCHAR(10), 12> last_name VARCHAR(10), 13> salary DECIMAL(10,2), 14> start_Date DATETIME, 15> region VARCHAR(10), 16> city VARCHAR(20), 17> managerid INTEGER 18> ); 19> 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> DECLARE 5> @ID int 6> SELECT 7> @ID = 0 8> SELECT 9> @ID = ID 10> FROM 11> Employee 12> WHERE 13> ID = 1 14> SELECT 15> @ID 16> 17> 18> drop table employee; 19> GO


         1

(1 rows affected) 1></source>


multiple variables can be assigned values in a single operation.

   <source lang="sql">

6> 7> DECLARE @MyNumber1 Int, @MyNumber2 Int, @MyResult1 Int, @MyResult2 Int 8> SELECT @MyNumber1 = 144, @MyNumber2 = 121 9> -- Assign the function result to the variable: 10> SELECT @MyResult1 = SQRT(@MyNumber1), @MyResult2 = SQRT(@MyNumber2) 11> -- Return the variable value 12> SELECT @MyResult1, @MyResult2 13> GO


-----------
        12          11

(1 rows affected)</source>


Syntax for Variable Assignment with the SELECT Statement

   <source lang="sql">

SELECT {@local_variable = expression } [,...n]</source>


Use select to output a variable

   <source lang="sql">

3> CREATE TABLE authors( 4> au_id varchar(11), 5> au_lname varchar(40) NOT NULL, 6> au_fname varchar(20) NOT NULL, 7> phone char(12) NOT NULL DEFAULT ("UNKNOWN"), 8> address varchar(40) NULL, 9> city varchar(20) NULL, 10> state char(2) NULL, 11> zip char(5) NULL, 12> contract bit NOT NULL 13> ) 14> GO 1> insert authors values("1", "Joe", "Abra", "111 111-1111", "6 St.", "Berkeley", "CA", "11111", 1) 2> insert authors values("2", "Jack", "Majo", "222 222-2222", "3 St.", "Oakland" , "CA", "22222", 1) 3> insert authors values("3", "Pink", "Cherry", "333 333-3333", "5 Ln.", "Vancouver", "BC", "33333", 1) 4> insert authors values("4", "Blue", "Albert", "444 444-4444", "7 Av.", "Vancouver", "BC", "44444", 1) 5> insert authors values("5", "Red", "Anne", "555 555-5555", "6 Av.", "Regina", "SK", "55555", 1) 6> insert authors values("6", "Black", "Michel", "666 666-6666", "3 Pl.", "Regina", "SK", "66666", 1) 7> insert authors values("7", "White", "Sylvia", "777 777-7777", "1 Pl.", "Rockville", "MD", "77777", 1) 8> insert authors values("8", "Yellow","Heather","888 888-8888", "3 Pu", "Vacaville", "CA", "88888", 0) 9> insert authors values("9", "Gold", "Dep", "999 999-9999", "5 Av.", "Oakland", "CA", "99999", 0) 10> insert authors values("10", "Siler", "Dean", "000 000-0000", "4 Av.", "Oakland", "CA", "00000", 1) 11> GO (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) 1> 2> 3> DECLARE @firstname varchar(20) 4> SELECT @firstname = au_fname 5> FROM authors 6> WHERE au_lname = "Greene" 7> 8> SELECT @firstname 9> SELECT @firstname = au_fname 10> FROM authors 11> WHERE au_lname = "Ben-Gan" 12> SELECT @firstname 13> GO


NULL (1 rows affected)


NULL (1 rows affected) 1> 2> drop table authors; 3> GO</source>


Using SELECT to Assign Variables

   <source lang="sql">

3> DECLARE @MyNumber1 Int, @MyNumber2 Int, @MyResult1 Int, @MyResult2 Int 4> SELECT @MyNumber1 = 144, @MyNumber2 = 121 5> 6> SELECT @MyResult1 = SQRT(@MyNumber1), @MyResult2 = SQRT(@MyNumber2) 7> 8> SELECT @MyResult1, @MyResult2 9> GO


-----------
        12          11

(1 rows affected)</source>