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» 《SQL Anywhere Studio 9开发指南》代码
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[参考文档]
《SQL Anywhere Studio 9开发指南》代码
ninetailsfox
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#31
使用道具
发表于 2007-11-24 23:05
3.23 CREATE VIEW [Top]
-----------------------------------------------------------------------------------
CREATE VIEW v_parent_child AS
SELECT USER_NAME ( parent_table.creator ) AS parent_owner,
parent_table.table_name AS parent_table,
USER_NAME ( child_table.creator ) AS child_owner,
child_table.table_name AS child_table
FROM SYS.SYSFOREIGNKEY AS foreign_key
INNER JOIN
( SELECT table_id,
creator,
table_name
FROM SYS.SYSTABLE
WHERE table_type = 'BASE' ) -- no VIEWs, etc.
AS parent_table
ON parent_table.table_id = foreign_key.primary_table_id
INNER JOIN
( SELECT table_id,
creator,
table_name
FROM SYS.SYSTABLE
WHERE table_type = 'BASE' ) -- no VIEWs, etc.
AS child_table
ON child_table.table_id = foreign_key.foreign_table_id;
-----------------------------------------------------------------------------------
SELECT parent_owner,
parent_table,
child_owner,
child_table
FROM v_parent_child
WHERE parent_owner = 'DBA'
AND child_owner = 'DBA'
ORDER BY 1, 2, 3, 4;
-----------------------------------------------------------------------------------
CREATE TABLE parent (
key_1 INTEGER NOT NULL PRIMARY KEY,
non_key_1 INTEGER NOT NULL );
CREATE VIEW v_parent AS
SELECT *
FROM parent;
CREATE TABLE child (
key_1 INTEGER NOT NULL REFERENCES parent ( key_1 ),
key_2 INTEGER NOT NULL,
non_key_1 INTEGER NOT NULL,
PRIMARY KEY ( key_1, key_2 ) );
CREATE VIEW v_child AS
SELECT *
FROM child;
CREATE VIEW v_family (
parent_key_1,
parent_non_key_1,
child_key_1,
child_key_2,
child_non_key_1 ) AS
SELECT parent.key_1,
parent.non_key_1,
child.key_1,
child.key_2,
child.non_key_1
FROM parent
INNER JOIN child
ON child.key_1 = parent.key_1;
INSERT v_parent VALUES ( 1, 444 );
INSERT v_parent VALUES ( 2, 555 );
INSERT v_parent VALUES ( 3, 666 );
INSERT v_child VALUES ( 1, 77, 777 );
INSERT v_child VALUES ( 1, 88, 888 );
INSERT v_child VALUES ( 2, 99, 999 );
INSERT v_child VALUES ( 3, 11, 111 );
UPDATE v_family
SET parent_non_key_1 = 1111,
child_non_key_1 = 2222
WHERE parent_key_1 = 1
AND child_key_2 = 88;
DELETE v_child
WHERE key_1 = 3
AND key_2 = 11;
SELECT * FROM v_family
ORDER BY parent_key_1,
child_key_2;
-----------------------------------------------------------------------------------
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ninetailsfox
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论坛徽章:2
#32
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发表于 2007-11-27 08:16
3.24 WITH Clause [Top]
-----------------------------------------------------------------------------------
CREATE VIEW v_parent_child AS
WITH base_table AS
( SELECT table_id,
creator,
table_name
FROM SYS.SYSTABLE
WHERE table_type = 'BASE' )
SELECT USER_NAME ( parent_table.creator ) AS parent_owner,
parent_table.table_name AS parent_table,
USER_NAME ( child_table.creator ) AS child_owner,
child_table.table_name AS child_table
FROM SYS.SYSFOREIGNKEY AS foreign_key
INNER JOIN base_table
AS parent_table
ON parent_table.table_id = foreign_key.primary_table_id
INNER JOIN base_table
AS child_table
ON child_table.table_id = foreign_key.foreign_table_id;
-----------------------------------------------------------------------------------
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论坛徽章:2
#33
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发表于 2007-11-27 08:16
3.24.1 Recursive UNION [Top]
-----------------------------------------------------------------------------------
CREATE TABLE employee (
employee_id INTEGER NOT NULL,
manager_id INTEGER NOT NULL REFERENCES employee ( employee_id ),
name VARCHAR ( 20 ) NOT NULL,
salary NUMERIC ( 20, 2 ) NOT NULL,
PRIMARY KEY ( employee_id ) );
INSERT INTO employee VALUES ( 1, 1, 'Ainslie', 1000000.00 );
INSERT INTO employee VALUES ( 2, 1, 'Briana', 900000.00 );
INSERT INTO employee VALUES ( 3, 1, 'Calista', 900000.00 );
INSERT INTO employee VALUES ( 4, 1, 'Delmar', 900000.00 );
INSERT INTO employee VALUES ( 5, 2, 'Electra', 750000.00 );
INSERT INTO employee VALUES ( 6, 3, 'Fabriane', 800000.00 );
INSERT INTO employee VALUES ( 7, 3, 'Genevieve', 750000.00 );
INSERT INTO employee VALUES ( 8, 4, 'Hunter', 800000.00 );
INSERT INTO employee VALUES ( 9, 6, 'Inari', 500000.00 );
INSERT INTO employee VALUES ( 10, 6, 'Jordan', 100000.00 );
INSERT INTO employee VALUES ( 11, 8, 'Khalil', 100000.00 );
INSERT INTO employee VALUES ( 12, 8, 'Lisette', 100000.00 );
INSERT INTO employee VALUES ( 13, 10, 'Marlon', 100000.00 );
INSERT INTO employee VALUES ( 14, 10, 'Nissa', 100000.00 );
-----------------------------------------------------------------------------------
WITH RECURSIVE superior_list
( level,
chosen_employee_id,
manager_id,
employee_id,
name )
AS ( SELECT CAST ( 1 AS INTEGER ) AS level,
employee.employee_id AS chosen_employee_id,
employee.manager_id AS manager_id,
employee.employee_id AS employee_id,
employee.name AS name
FROM employee
UNION ALL
SELECT superior_list.level + 1,
superior_list.chosen_employee_id,
employee.manager_id,
employee.employee_id,
employee.name
FROM superior_list
INNER JOIN employee
ON employee.employee_id = superior_list.manager_id
WHERE superior_list.level <= 99
AND superior_list.manager_id <> superior_list.employee_id )
SELECT superior_list.level,
superior_list.name
FROM superior_list
WHERE superior_list.chosen_employee_id = 13
ORDER BY superior_list.level DESC;
-----------------------------------------------------------------------------------
CREATE VIEW v_superior_list AS
WITH RECURSIVE superior_list
( level,
chosen_employee_id,
manager_id,
employee_id,
name )
AS ( SELECT CAST ( 1 AS INTEGER ) AS level,
employee.employee_id AS chosen_employee_id,
employee.manager_id AS manager_id,
employee.employee_id AS employee_id,
employee.name AS name
FROM employee
UNION ALL
SELECT superior_list.level + 1,
superior_list.chosen_employee_id,
employee.manager_id,
employee.employee_id,
employee.name
FROM superior_list
INNER JOIN employee
ON employee.employee_id = superior_list.manager_id
WHERE superior_list.level <= 99
AND superior_list.manager_id <> superior_list.employee_id )
SELECT *
FROM superior_list;
-----------------------------------------------------------------------------------
SELECT v_superior_list.level,
v_superior_list.name
FROM v_superior_list
WHERE v_superior_list.chosen_employee_id = 13
ORDER BY v_superior_list.level DESC;
-----------------------------------------------------------------------------------
SELECT LIST ( v_superior_list.name,
', then '
ORDER BY v_superior_list.level ASC ) AS "Khalil's Superiors"
FROM v_superior_list
WHERE v_superior_list.chosen_employee_id = 11
AND v_superior_list.level > 1;
-----------------------------------------------------------------------------------
CREATE VIEW v_salary_list AS
WITH RECURSIVE salary_list
( level,
chosen_employee_id,
manager_id,
employee_id,
name,
salary )
AS ( SELECT CAST ( 1 AS INTEGER ) AS level,
employee.employee_id AS chosen_employee_id,
employee.manager_id AS manager_id,
employee.employee_id AS employee_id,
employee.name AS name,
employee.salary AS salary
FROM employee
UNION ALL
SELECT salary_list.level + 1,
salary_list.chosen_employee_id,
employee.manager_id,
employee.employee_id,
employee.name,
employee.salary
FROM salary_list
INNER JOIN employee
ON employee.manager_id = salary_list.employee_id
WHERE salary_list.level <= 99
AND employee.manager_id <> employee.employee_id )
SELECT *
FROM salary_list;
-----------------------------------------------------------------------------------
SELECT employee.name,
( SELECT SUM ( v_salary_list.salary )
FROM v_salary_list
WHERE v_salary_list.chosen_employee_id
= employee.employee_id ) AS payroll
FROM employee
ORDER BY 1;
-----------------------------------------------------------------------------------
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ninetailsfox
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论坛徽章:2
#34
使用道具
发表于 2007-11-27 08:16
3.25 UNLOAD TABLE and UNLOAD SELECT [Top]
-----------------------------------------------------------------------------------
UNLOAD TABLE t1 TO 't1_a1.txt';
UNLOAD SELECT * FROM t1 TO 't1_a2.txt';
-----------------------------------------------------------------------------------
CREATE TABLE t1 (
key_1 INTEGER NOT NULL,
col_2 VARCHAR ( 100 ) NULL,
col_3 BINARY ( 100 ) NULL,
col_4 DECIMAL ( 11, 2 ) NULL,
col_5 DATE NULL,
col_6 INTEGER NOT NULL,
PRIMARY KEY ( key_1 ) );
INSERT t1 VALUES (
1, 'Fred''s Here', 'Fred''s Here', 12.34, '2003-09-30', 888 );
INSERT t1 VALUES (
2, 0x74776f0d0a6c696e6573, 'two\x0d\x0alines', 67.89, '2003-09-30', 999 );
COMMIT;
UNLOAD SELECT * FROM t1 ORDER BY key_1
TO 't1_b1.txt';
UNLOAD SELECT * FROM t1 ORDER BY key_1
TO 't1_b2.txt' ESCAPES OFF;
UNLOAD SELECT * FROM t1 ORDER BY key_1
TO 't1_b3.txt' ESCAPES OFF QUOTES OFF;
UNLOAD SELECT * FROM t1 ORDER BY key_1
TO 't1_b4.txt' HEXADECIMAL OFF ESCAPES OFF QUOTES OFF;
UNLOAD SELECT * FROM t1 ORDER BY key_1
TO 't1_b5.txt' DELIMITED BY '' HEXADECIMAL OFF ESCAPES OFF QUOTES OFF;
-----------------------------------------------------------------------------------
SELECT STRING (
key_1,
col_2,
col_3,
col_4,
col_5,
col_6 )
FROM t1
ORDER BY key_1;
-----------------------------------------------------------------------------------
UNLOAD
SELECT STRING (
'<HTML><BODY><OL>\x0d\x0a',
' <LI>',
LIST ( DISTINCT state,
'</LI>\x0d\x0a <LI>'
ORDER BY state ),
'</LI>\x0d\x0a',
'</OL></BODY></HTML>' ) AS states_page
FROM employee
WHERE dept_id = 100
TO 'c:\\temp\\states_page.html' ESCAPES OFF QUOTES OFF;
-----------------------------------------------------------------------------------
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ninetailsfox
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论坛徽章:2
#35
使用道具
发表于 2007-11-27 08:16
3.26 ISQL OUTPUT [Top]
-----------------------------------------------------------------------------------
SELECT *
FROM product
WHERE name = 'Sweatshirt'
ORDER BY id;
OUTPUT TO 'product.txt';
-----------------------------------------------------------------------------------
SELECT id,
name,
quantity,
unit_price
FROM product
WHERE name = 'Sweatshirt'
ORDER BY id;
OUTPUT TO 'c:\\temp\\product.dta' COLUMN WIDTHS ( 5, 15, 5, 7 ) FORMAT FIXED;
OUTPUT TO 'c:\\temp\\product.html' FORMAT HTML;
OUTPUT TO 'c:\\temp\\product.xml' FORMAT XML;
-----------------------------------------------------------------------------------
CREATE TABLE t1 (
key_1 INTEGER NOT NULL,
col_2 VARCHAR ( 100 ) NULL,
col_3 BINARY ( 100 ) NULL,
col_4 DECIMAL ( 11, 2 ) NULL,
col_5 DATE NULL,
col_6 INTEGER NOT NULL,
PRIMARY KEY ( key_1 ) );
INSERT t1 VALUES (
1, 'Fred''s Here', 'Fred''s Here', 12.34, '2003-09-30', 888 );
INSERT t1 VALUES (
2, 0x74776f0d0a6c696e6573, 'two\x0d\x0alines', 67.89, '2003-09-30', 999 );
COMMIT;
SELECT * FROM t1 ORDER BY key_1;
OUTPUT TO 't1_c1.txt';
OUTPUT TO 't1_c2.txt' DELIMITED BY '' HEXADECIMAL ASIS QUOTE '';
-----------------------------------------------------------------------------------
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ninetailsfox
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论坛徽章:2
#36
使用道具
发表于 2007-11-27 08:17
3.27 Chapter Summary [Top]
-----------------------------------------------------------------------------------
Chapter 4 - UPDATING [Top]
-----------------------------------------------------------------------------------
4.1 Introduction [Top]
-----------------------------------------------------------------------------------
4.2 Single-Row UPDATE [Top]
-----------------------------------------------------------------------------------
UPDATE customer
SET address = '114 PIONEER AV'
WHERE id = 101;
-----------------------------------------------------------------------------------
UPDATE customer
SET address = '114 PIONEER AV'
WHERE id = 101
AND address = '114 Pioneer Avenue';
-----------------------------------------------------------------------------------
UPDATE customer
SET fname = 'Fred',
lname = 'Jones'
WHERE company_name = 'The Power Group';
-----------------------------------------------------------------------------------
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ninetailsfox
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论坛徽章:2
#37
使用道具
发表于 2007-11-27 08:17
4.3 Multi-Row UPDATE [Top]
-----------------------------------------------------------------------------------
UPDATE employee
SET salary = salary * 1.05;
-----------------------------------------------------------------------------------
UPDATE employee
SET termination_date = CURRENT DATE
WHERE dept_id = 100;
-----------------------------------------------------------------------------------
4.4 Logical Execution of a Set UPDATE [Top]
-----------------------------------------------------------------------------------
CREATE TABLE t1 (
key_1 UNSIGNED INTEGER NOT NULL PRIMARY KEY,
non_key_1 INTEGER NOT NULL );
CREATE TABLE t2 (
key_1 UNSIGNED INTEGER NOT NULL PRIMARY KEY,
non_key_1 INTEGER NOT NULL );
INSERT t1 VALUES ( 1, 0 );
INSERT t1 VALUES ( 2, 0 );
INSERT t2 VALUES ( 1, 100 );
INSERT t2 VALUES ( 2, 100 );
-----------------------------------------------------------------------------------
UPDATE t1
CROSS JOIN t1 AS x
CROSS JOIN t2
SET t1.non_key_1 = NUMBER(*),
t2.non_key_1 = NUMBER(*)
WHERE t1.non_key_1 = 0
ORDER BY t1.key_1 DESC,
t2.key_1 DESC;
-----------------------------------------------------------------------------------
-- Step 1:
SELECT ...
FROM t1
CROSS JOIN t1 AS x
CROSS JOIN t2
WHERE t1.non_key_1 = 0
ORDER BY t1.key_1 DESC,
t2.key_1 DESC;
-----------------------------------------------------------------------------------
-- Step 3
SELECT t1.*,
t2.*,
NUMBER(*)
FROM t1
CROSS JOIN t1 AS x
CROSS JOIN t2
WHERE t1.non_key_1 = 0
ORDER BY t1.key_1 DESC,
t2.key_1 DESC;
-----------------------------------------------------------------------------------
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ninetailsfox
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论坛徽章:2
#38
使用道具
发表于 2007-11-27 08:17
4.4.1 Set UPDATE [Top]
-----------------------------------------------------------------------------------
CREATE VIEW v1 AS
SELECT * FROM t1;
CREATE PROCEDURE p1()
BEGIN
SELECT * FROM t1;
END;
UPDATE v1
CROSS JOIN p1() AS x
CROSS JOIN ( SELECT * FROM t2 ) AS d2
SET v1.non_key_1 = NUMBER(*),
d2.non_key_1 = NUMBER(*)
WHERE v1.non_key_1 = 0
ORDER BY v1.key_1 DESC,
d2.key_1 DESC;
-----------------------------------------------------------------------------------
UPDATE employee
INNER JOIN ( SELECT FIRST sales_order.sales_rep AS top_rep_id
FROM sales_order
INNER JOIN sales_order_items
ON sales_order_items.id = sales_order.id
INNER JOIN product
ON product.id = sales_order_items.prod_id
GROUP BY sales_order.sales_rep
ORDER BY SUM ( sales_order_items.quantity
* product.unit_price ) DESC )
AS top_rep
ON top_rep.top_rep_id = employee.emp_id
SET employee.salary = employee.salary * 1.1;
-----------------------------------------------------------------------------------
UPDATE employee SET salary = 43230.00 WHERE emp_id = 299; -- salary was 39300.00
-----------------------------------------------------------------------------------
CREATE PROCEDURE p_top_salespeople
( IN @top_count INTEGER )
RESULT ( top_rep_id INTEGER )
BEGIN
DECLARE @select LONG VARCHAR;
SET @select = STRING (
'SELECT TOP ', @top_count, ' sales_order.sales_rep
FROM sales_order
INNER JOIN sales_order_items
ON sales_order_items.id = sales_order.id
INNER JOIN product
ON product.id = sales_order_items.prod_id
GROUP BY sales_order.sales_rep
ORDER BY SUM ( sales_order_items.quantity
* product.unit_price ) DESC' );
EXECUTE IMMEDIATE @select;
END;
UPDATE employee
INNER JOIN p_top_salespeople ( 3 )
ON p_top_salespeople.top_rep_id = employee.emp_id
SET employee.salary = employee.salary * 1.1;
-----------------------------------------------------------------------------------
UPDATE TOP 2
employee
INNER JOIN department
ON department.dept_id = employee.dept_id
SET employee.salary = employee.salary * 0.95
WHERE department.dept_name = 'Finance'
ORDER BY employee.start_date DESC;
-----------------------------------------------------------------------------------
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ninetailsfox
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论坛徽章:2
#39
使用道具
发表于 2007-11-27 08:17
4.5 UPDATE WHERE CURRENT OF Cursor [Top]
-----------------------------------------------------------------------------------
BEGIN
DECLARE @t1_key_1 INTEGER;
DECLARE @t1_non_key_1 INTEGER;
DECLARE @t2_key_1 INTEGER;
DECLARE @t2_non_key_1 INTEGER;
DECLARE @number INTEGER;
DECLARE @SQLSTATE VARCHAR ( 5 );
DECLARE cloop1 CURSOR FOR
SELECT t1.key_1,
t1.non_key_1,
t2.key_1,
t2.non_key_1
FROM t1
CROSS JOIN t1 AS x
CROSS JOIN t2
WHERE t1.non_key_1 = 0
ORDER BY t1.key_1 DESC,
t2.key_1 DESC;
OPEN cloop1;
FETCH cloop1 INTO
@t1_key_1,
@t1_non_key_1,
@t2_key_1,
@t2_non_key_1;
SET @SQLSTATE = SQLSTATE;
SET @number = 0;
WHILE ( @SQLSTATE IN ( '00000', '01W04' ) ) LOOP
SET @number = @number + 1;
UPDATE t1, t2
SET t1.non_key_1 = @number,
t2.non_key_1 = @number
WHERE CURRENT OF cloop1;
FETCH cloop1 INTO
@t1_key_1,
@t1_non_key_1,
@t2_key_1,
@t2_non_key_1;
SET @SQLSTATE = SQLSTATE;
END LOOP;
CLOSE cloop1;
END;
-----------------------------------------------------------------------------------
4.6 Chapter Summary [Top]
-----------------------------------------------------------------------------------
Chapter 5 - DELETING [Top]
-----------------------------------------------------------------------------------
5.1 Introduction [Top]
-----------------------------------------------------------------------------------
5.2 Single-Row DELETE [Top]
-----------------------------------------------------------------------------------
DELETE sales_order_items
WHERE id = 2015
AND line_id = 4;
-----------------------------------------------------------------------------------
5.3 Multi-Row DELETE [Top]
-----------------------------------------------------------------------------------
DELETE sales_order_items;
-----------------------------------------------------------------------------------
DELETE sales_order_items
WHERE id = 2015;
-----------------------------------------------------------------------------------
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ninetailsfox
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论坛徽章:2
#40
使用道具
发表于 2007-11-27 08:17
5.4 Logical Execution of a Set DELETE [Top]
-----------------------------------------------------------------------------------
CREATE TABLE t1 (
key_1 UNSIGNED INTEGER NOT NULL PRIMARY KEY,
non_key_1 INTEGER NOT NULL );
INSERT t1 VALUES ( 1, 1 );
INSERT t1 VALUES ( 2, 2 );
INSERT t1 VALUES ( 3, 3 );
INSERT t1 VALUES ( 4, 4 );
INSERT t1 VALUES ( 5, 5 );
-----------------------------------------------------------------------------------
DELETE t1
FROM t1
CROSS JOIN t1 AS x
WHERE t1.key_1 = 2;
-----------------------------------------------------------------------------------
-- Step 1
SELECT DISTINCT ...
FROM t1
CROSS JOIN t1 AS x
WHERE t1.key_1 = 2;
-----------------------------------------------------------------------------------
-- Step 3
SELECT DISTINCT t1.*
FROM t1
CROSS JOIN t1 AS x
WHERE t1.key_1 = 2;
-----------------------------------------------------------------------------------
-- Step 5
DELETE t1 WHERE key_1 = 2;
-----------------------------------------------------------------------------------
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