How to use XMLAGG SQL Functions ..
SQL function XMLAgg is an aggregate function that produces a forest of XML elements from a collection of XML elements.
Example 1:
Generating Department Elements with a List of Employee Elements
This example produces a Department element containing Employee elements with employee job ID and last name as the contents of the elements. It also orders the employee XML elements in the department by their last name. (The actual result is not pretty-printed.)
SELECT XMLElement("Department", XMLAgg(XMLElement("Employee",
e.job_id||' '||e.last_name)
ORDER BY e.last_name))
AS "Dept_list"
FROM hr.employees e
WHERE e.department_id = 30 OR e.department_id = 40;
Example 2:
The result is a single row, because XMLAgg aggregates the rows. You can use the GROUP BY clause to group the returned set of rows into multiple groups. (The actual result of the following query is not pretty-printed.)
SELECT XMLElement("Department", XMLAttributes(department_id AS "deptno"),
XMLAgg(XMLElement("Employee", e.job_id||' '||e.last_name)))
AS "Dept_list"
FROM hr.employees e
GROUP BY e.department_id;
Example 3:
XMLAGG: Generating Nested Elements
Function XMLAgg can be used to reflect the hierarchical nature of some relationships that exist in tables. This example generates a department element for department 30. Within this element is a child element for each employee of the department. Within each employee element is a dependent element for each dependent of that employee.
First, this query shows the employees of department 30.
SELECT last_name, employee_id FROM employees WHERE department_id = 30;
LAST_NAME EMPLOYEE_ID
------------------------- -----------
Raphaely 114
Khoo 115
Baida 116
Tobias 117
Himuro 118
Colmenares 119
6 rows selected.
A dependents table is created, to hold the dependents of each employee.
CREATE TABLE hr.dependents (id NUMBER(4) PRIMARY KEY,
employee_id NUMBER(4),
name VARCHAR2(10));
Table created.
INSERT INTO dependents VALUES (1, 114, 'MARK');
1 row created.
INSERT INTO dependents VALUES (2, 114, 'JACK');
1 row created.
INSERT INTO dependents VALUES (3, 115, 'JANE');
1 row created.
INSERT INTO dependents VALUES (4, 116, 'HELEN');
1 row created.
INSERT INTO dependents VALUES (5, 116, 'FRANK');
1 row created.
COMMIT;
Commit complete.
This query generates the XML data for department that contains the information on dependents (the actual output is not pretty-printed):
SELECT
XMLElement(
"Department",
XMLAttributes(d.department_name AS "name"),
(SELECT
XMLAgg(XMLElement("emp",
XMLAttributes(e.last_name AS name),
(SELECT XMLAgg(XMLElement("dependent",
XMLAttributes(de.name AS "name")))
FROM dependents de
WHERE de.employee_id = e.employee_id)))
FROM employees e
WHERE e.department_id = d.department_id)) AS "dept_list"
FROM departments d
WHERE department_id = 30;
output
--------------------------------------------------------------------------------
dept_list
--------------------------------------------------------------------------------
.
1 row selected in xml format it not showing in blog due to xml format ...
SQL function XMLAgg is an aggregate function that produces a forest of XML elements from a collection of XML elements.
Example 1:
Generating Department Elements with a List of Employee Elements
This example produces a Department element containing Employee elements with employee job ID and last name as the contents of the elements. It also orders the employee XML elements in the department by their last name. (The actual result is not pretty-printed.)
SELECT XMLElement("Department", XMLAgg(XMLElement("Employee",
e.job_id||' '||e.last_name)
ORDER BY e.last_name))
AS "Dept_list"
FROM hr.employees e
WHERE e.department_id = 30 OR e.department_id = 40;
Example 2:
The result is a single row, because XMLAgg aggregates the rows. You can use the GROUP BY clause to group the returned set of rows into multiple groups. (The actual result of the following query is not pretty-printed.)
SELECT XMLElement("Department", XMLAttributes(department_id AS "deptno"),
XMLAgg(XMLElement("Employee", e.job_id||' '||e.last_name)))
AS "Dept_list"
FROM hr.employees e
GROUP BY e.department_id;
Example 3:
XMLAGG: Generating Nested Elements
Function XMLAgg can be used to reflect the hierarchical nature of some relationships that exist in tables. This example generates a department element for department 30. Within this element is a child element for each employee of the department. Within each employee element is a dependent element for each dependent of that employee.
First, this query shows the employees of department 30.
SELECT last_name, employee_id FROM employees WHERE department_id = 30;
LAST_NAME EMPLOYEE_ID
------------------------- -----------
Raphaely 114
Khoo 115
Baida 116
Tobias 117
Himuro 118
Colmenares 119
6 rows selected.
A dependents table is created, to hold the dependents of each employee.
CREATE TABLE hr.dependents (id NUMBER(4) PRIMARY KEY,
employee_id NUMBER(4),
name VARCHAR2(10));
Table created.
INSERT INTO dependents VALUES (1, 114, 'MARK');
1 row created.
INSERT INTO dependents VALUES (2, 114, 'JACK');
1 row created.
INSERT INTO dependents VALUES (3, 115, 'JANE');
1 row created.
INSERT INTO dependents VALUES (4, 116, 'HELEN');
1 row created.
INSERT INTO dependents VALUES (5, 116, 'FRANK');
1 row created.
COMMIT;
Commit complete.
This query generates the XML data for department that contains the information on dependents (the actual output is not pretty-printed):
SELECT
XMLElement(
"Department",
XMLAttributes(d.department_name AS "name"),
(SELECT
XMLAgg(XMLElement("emp",
XMLAttributes(e.last_name AS name),
(SELECT XMLAgg(XMLElement("dependent",
XMLAttributes(de.name AS "name")))
FROM dependents de
WHERE de.employee_id = e.employee_id)))
FROM employees e
WHERE e.department_id = d.department_id)) AS "dept_list"
FROM departments d
WHERE department_id = 30;
output
--------------------------------------------------------------------------------
dept_list
--------------------------------------------------------------------------------
.
1 row selected in xml format it not showing in blog due to xml format ...
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