SQL Server/T-SQL/Select Query/NULL — различия между версиями
Admin (обсуждение | вклад) м (1 версия) |
Admin (обсуждение | вклад) м (1 версия) |
(нет различий)
| |
Текущая версия на 10:21, 26 мая 2010
Count function deals with NULL values
1> create table employee(
2> ID int,
3> name nvarchar (10),
4> salary int,
5> start_date datetime,
6> city nvarchar (10),
7> region char (1))
8> GO
1>
2> insert into employee (ID, name, salary, start_date, city, region)
3> values (1, "Jason", 40420, "02/01/94", "New York", "W")
4> GO
(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (2, "Robert",14420, "01/02/95", "Vancouver","N")
3> GO
(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (3, "Celia", 24020, "12/03/96", "Toronto", "W")
3> GO
(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (4, "Linda", 40620, "11/04/97", "New York", "N")
3> GO
(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (5, "David", 80026, "10/05/98", "Vancouver","W")
3> GO
(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (6, "James", 70060, "09/06/99", "Toronto", "N")
3> GO
(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (7, "Alison",90620, "08/07/00", "New York", "W")
3> GO
(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (8, NULL, 26020, "07/08/01", "Vancouver","N")
3> GO
(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (9, NULL, 60020, "06/09/02", "Toronto", "W")
3> GO
(1 rows affected)
1>
2> select * from employee
3> GO
ID name salary start_date city region
----------- ---------- ----------- ----------------------- ---------- ------
1 Jason 40420 1994-02-01 00:00:00.000 New York W
2 Robert 14420 1995-01-02 00:00:00.000 Vancouver N
3 Celia 24020 1996-12-03 00:00:00.000 Toronto W
4 Linda 40620 1997-11-04 00:00:00.000 New York N
5 David 80026 1998-10-05 00:00:00.000 Vancouver W
6 James 70060 1999-09-06 00:00:00.000 Toronto N
7 Alison 90620 2000-08-07 00:00:00.000 New York W
8 NULL 26020 2001-07-08 00:00:00.000 Vancouver N
9 NULL 60020 2002-06-09 00:00:00.000 Toronto W
(9 rows affected)
1>
2>
3> --Count does not eliminate duplicate or NULL values.
4> select count(name) from employee
5>
6>
7>
8>
9>
10> drop table employee
11> GO
-----------
7
Warning: Null value is eliminated by an aggregate or other SET operation.
1>
NOT NULL column
13>
14> CREATE TABLE project (project_no CHAR(4) NOT NULL,
15> project_name CHAR(15) NOT NULL,
16> budget FLOAT NULL)
17>
18> insert into project values ("p1", "Search Engine", 120000.00)
19> insert into project values ("p2", "Programming", 95000.00)
20> insert into project values ("p3", "SQL", 186500.00)
21>
22>
23>
24> -- VAR computes the variance of all the values listed in a column or expression
25>
26> SELECT VAR(budget) variance_of_budgets FROM project
27>
28> drop table project
29> GO
(1 rows affected)
(1 rows affected)
(1 rows affected)
variance_of_budgets
------------------------
2236583333.3333321
(1 rows affected)
1>
Replacing a NULL Value with an Alternative Value
4>
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> managerid INTEGER
14> );
15> 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" , NULL, 4234,"2007-07-17","West","Regina",9);
2> GO
(1 rows affected)
1> INSERT INTO employee VALUES (9, "Joan" , NULL, 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 NULL 4234.00 2007-07-17 00:00:00.000 West Regina 9
9 Joan NULL 6123.00 2001-04-16 00:00:00.000 North Toronto 10
(9 rows affected)
1>
2>
3>
4> SELECT Last_Name,
5> ISNULL(last_name, "UNKNOWN") Title
6> FROM employee
7> GO
Last_Name Title
---------- ----------
Martin Martin
Mathews Mathews
Smith Smith
Rice Rice
Black Black
Green Green
Larry Larry
NULL UNKNOWN
NULL UNKNOWN
(9 rows affected)
1>
2> drop table employee;
3> GO