SQL Server/T-SQL/Constraints/Maintain Constaints — различия между версиями

Материал из SQL эксперт
Перейти к: навигация, поиск
м (1 версия)
 
м (1 версия)
 
(нет различий)

Текущая версия на 10:21, 26 мая 2010

Maintenance of the referential constraint

28>
29> CREATE TABLE works_on       (emp_no       INTEGER NOT NULL,
30>                         project_no    CHAR(4) NOT NULL,
31>                         job CHAR (15) NULL,
32>                         enter_date    DATETIME NULL)
33>
34> insert into works_on values (1, "p1", "analyst", "1997.10.1")
35> insert into works_on values (1, "p3", "manager", "1999.1.1")
36> insert into works_on values (2, "p2", "clerk",   "1998.2.15")
37> insert into works_on values (2, "p2",  NULL,     "1998.6.1")
38> insert into works_on values (3, "p2",  NULL,     "1997.12.15")
39> insert into works_on values (4, "p3", "analyst", "1998.10.15")
40> insert into works_on values (5, "p1", "manager", "1998.4.15")
41> insert into works_on values (6, "p1",  NULL,     "1998.8.1")
42> insert into works_on values (7, "p2", "clerk",   "1999.2.1")
43> insert into works_on values (8, "p3", "clerk",   "1997.11.15")
44> insert into works_on values (7, "p1", "clerk",   "1998.1.4")
45>
46> select * from works_on
47> 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 rows affected)
emp_no      project_no job             enter_date
----------- ---------- --------------- -----------------------
          1 p1         analyst         1997-10-01 00:00:00.000
          1 p3         manager         1999-01-01 00:00:00.000
          2 p2         clerk           1998-02-15 00:00:00.000
          2 p2         NULL            1998-06-01 00:00:00.000
          3 p2         NULL            1997-12-15 00:00:00.000
          4 p3         analyst         1998-10-15 00:00:00.000
          5 p1         manager         1998-04-15 00:00:00.000
          6 p1         NULL            1998-08-01 00:00:00.000
          7 p2         clerk           1999-02-01 00:00:00.000
          8 p3         clerk           1997-11-15 00:00:00.000
          7 p1         clerk           1998-01-04 00:00:00.000
(11 rows affected)
1> CREATE TABLE employee  (emp_no    INTEGER NOT NULL,
2>                         emp_fname CHAR(20) NOT NULL,
3>                         emp_lname CHAR(20) NOT NULL,
4>                         dept_no   CHAR(4) NULL)
5>
6> insert into employee values(1,  "Matthew", "Smith",    "d3")
7> insert into employee values(2,  "Ann",     "Jones",    "d3")
8> insert into employee values(3,  "John",    "Barrimore","d1")
9> insert into employee values(4,  "James",   "James",    "d2")
10> insert into employee values(5,  "Elsa",    "Bertoni",  "d2")
11> insert into employee values(6,  "Elke",    "Hansel",   "d2")
12> insert into employee values(7,  "Sybill",  "Moser",    "d1")
13>
14> select * from employee
15> GO
(1 rows affected)
(1 rows affected)
(1 rows affected)
(1 rows affected)
(1 rows affected)
(1 rows affected)
(1 rows affected)
emp_no      emp_fname            emp_lname            dept_no
----------- -------------------- -------------------- -------
          1 Matthew              Smith                d3
          2 Ann                  Jones                d3
          3 John                 Barrimore            d1
          4 James                James                d2
          5 Elsa                 Bertoni              d2
          6 Elke                 Hansel               d2
          7 Sybill               Moser                d1
(7 rows affected)
1>
2> -- Maintenance of the referential constraint
3>
4> CREATE PROCEDURE delete_emp @employee_no INT, @counter INT OUTPUT
5> AS SELECT @counter = COUNT(*) FROM works_on WHERE emp_no = @employee_no
6>    DELETE FROM employee WHERE emp_no = @employee_no
7>    DELETE FROM works_on WHERE emp_no = @employee_no
8> GO
1>
2> DECLARE @C int
3> EXEC delete_emp 1, @C
4> Print @C
5> GO
(1 rows affected)
(2 rows affected)
(2 rows affected)
1>
2> select * from employee
3> select * from works_on
4> GO
emp_no      emp_fname            emp_lname            dept_no
----------- -------------------- -------------------- -------
          2 Ann                  Jones                d3
          3 John                 Barrimore            d1
          4 James                James                d2
          5 Elsa                 Bertoni              d2
          6 Elke                 Hansel               d2
          7 Sybill               Moser                d1
(6 rows affected)
emp_no      project_no job             enter_date
----------- ---------- --------------- -----------------------
          2 p2         clerk           1998-02-15 00:00:00.000
          2 p2         NULL            1998-06-01 00:00:00.000
          3 p2         NULL            1997-12-15 00:00:00.000
          4 p3         analyst         1998-10-15 00:00:00.000
          5 p1         manager         1998-04-15 00:00:00.000
          6 p1         NULL            1998-08-01 00:00:00.000
          7 p2         clerk           1999-02-01 00:00:00.000
          8 p3         clerk           1997-11-15 00:00:00.000
          7 p1         clerk           1998-01-04 00:00:00.000
(9 rows affected)
1>
2>
3> drop procedure delete_emp
4> drop table employee
5> drop table works_on
6> GO
1>
2>