|
Basic Syntax Elements
|
START WITH <condition>
CONNECT BY
[NOCYCLE]
<condition>
|
|
|
|
CONNECT BY PRIOR
|
|
A condition that identifies the relationship between parent rows and child rows of the hierarchy
|
CONNECT BY <child_value> = <parent_value>
|
conn hr/hr
SELECT employee_id, last_name, manager_id
FROM employees
CONNECT BY PRIOR employee_id = manager_id;
|
|
|
|
START WITH
|
|
Specifies a condition that identifies the row(s) to be used as the root(s) of a hierarchical query
|
START WITH (column_name) = <value>
|
SELECT last_name, employee_id, manager_id, LEVEL
FROM employees
START WITH employee_id = 100
CONNECT BY PRIOR employee_id = manager_id;
|
|
|
|
ORDER SIBLINGS BY
|
|
SIBLINGS BY preserves any ordering specified in the hierarchical query clause and then applies the order_by_clause to the siblings of the hierarchy
|
ORDER SIBLINGS BY (column_name)
|
SELECT last_name, employee_id, manager_id, LEVEL
FROM employees
START WITH employee_id = 100
CONNECT BY PRIOR employee_id = manager_id
ORDER SIBLINGS BY last_name;
|
|
|
|
CONNECT_BY_ROOT
|
CONNECT_BY_ROOT is a unary operator that is valid only in hierarchical queries. When you qualify a column with this operator, Oracle returns the column value using data from the root row.
Cannot be specified with the START WITH or CONNECT BY condition.
|
The following example returns the last name of each employee in department 110, each manager above that employee in the hierarchy, the number of levels between manager and employee, and the path between the two:
|
col emp format a20
col mgr format a20
set linesize 120
SELECT "Name", SUM(salary) "Total_Salary"
FROM (
SELECT CONNECT_BY_ROOT last_name "Name", salary
FROM employees
WHERE department_id = 110
CONNECT BY PRIOR employee_id = manager_id)
GROUP BY "Name";
-- Thanks Colin for the correction
|
|
|
|
CONNECT_BY_ISCYCLE Pseudocolumn
|
|
The CONNECT_BY_ISCYCLE pseudocolumn returns 1 if the current row has a child which is also its ancestor. Otherwise it returns 0
|
UPDATE employees SET manager_id = 145
WHERE employee_id = 100;
SELECT last_name, LEVEL, SYS_CONNECT_BY_PATH(last_name, '/') "Path"
FROM employees
WHERE LEVEL <= 3 AND department_id = 80
START WITH last_name = 'King'
CONNECT BY PRIOR employee_id = manager_id AND LEVEL <= 4;
2 3 4 5 6 7 ERROR:
ORA-01436: CONNECT BY loop in user data
SELECT last_name, CONNECT_BY_ISCYCLE "Cycle", LEVEL, SYS_CONNECT_BY_PATH(last_name, '/') "Path"
FROM employees
WHERE LEVEL <= 3 AND department_id = 80
START WITH last_name = 'King'
CONNECT BY NOCYCLE PRIOR employee_id = manager_id
AND LEVEL <= 4;
|
|
|
|
CONNECT_BY_ISLEAF Pseudocolumn
|
|
The CONNECT_BY_ISLEAF pseudocolumn returns 1 if the current row is a leaf of the tree defined by the CONNECT BY condition. Otherwise it returns 0. This information indicates whether a given row can be further expanded to show more of the hierarchy.
|
SELECT last_name "Employee", CONNECT_BY_ISLEAF "IsLeaf",
LEVEL, SYS_CONNECT_BY_PATH(last_name, '/') "Path"
FROM employees
WHERE level <= 3
AND department_id = 80
START WITH last_name = 'King'
CONNECT BY PRIOR employee_id = manager_id
AND LEVEL <= 4;
|
|
|
|
LEVEL Pseudocolumn
|
|
For each row returned by a hierarchical query, the LEVEL pseudocolumn returns 1 for a root row, 2 for a child of a root, and so on
|
SELECT employee_id, last_name, manager_id, LEVEL
FROM employees
CONNECT BY PRIOR employee_id = manager_id;
SELECT LPAD(' ',2*(LEVEL-1)) || last_name ORG_CHART,
employee_id, manager_id, job_id
FROM employees
START WITH job_id = 'AD_VP'
CONNECT BY PRIOR employee_id = manager_id;
|
|
|
|
SYS_CONNECT_BY_PATH
|
|
Returns the path of a column value from root to node, with column values separated by char for each row returned by CONNECT BY condition
|
|
SYS_CONNECT_BY_PATH(<column>, <char>)
|
|
See CONNECT_BY_ISCYCLE demo
|
|
|
|
Function Demo
|
|
Use A Function To Receive The Current Node and Search for Parents of the Current Node
|
CREATE OR REPLACE FUNCTION permissions_sub_tree_root (
the_id IN NUMBER,
the_level IN NUMBER)
RETURN NUMBER IS
sub_tree_root NUMBER(10);
BEGIN
SELECT id
INTO sub_tree_root
FROM hierarchy
WHERE level = the_level
-- Connect 'upwards', i.e. find the parent
CONNECT BY PRIOR PARENT = id
START WITH ID = the_id;
RETURN sub_tree_root;
END permissions_sub_tree_root;
/
SELECT id, name, username
FROM (
SELECT ID, PARENT, NAME,
permissions_sub_tree_root(id, LEVEL) ROOT
FROM hierarchy
CONNECT BY PRIOR id = PARENT) HIERARCHY, permissions
WHERE ROOT = hierarchy_id;
|
|
|
|
GROUP BY Demo
|
Group By Demo with CONNECT_BY_ROOT and
CONNECT_BY_PRIOR
|
SELECT name, SUM(salary) "Total_Salary"
FROM (
SELECT CONNECT_BY_ROOT last_name "Name", salary
FROM employees
WHERE department_id = 110
CONNECT BY PRIOR employee_id = manager_id)
GROUP BY name;
|
|
|
|
Demos
|
|
Indenting
|
col lname format a30
SELECT LPAD(' ', level*2, ' ') || last_name LNAME, d.department_id
FROM employees e, departments d
WHERE e.department_id = d.department_id
START WITH employee_id = 100
CONNECT BY PRIOR e.employee_id = e.manager_id;
|
|
Hierarchical Query with IN
|
In a [NOT] IN condition in a WHERE clause, if the right-hand side of the condition is a subquery, you cannot use LEVEL on the left-hand side of the condition. However, you can specify LEVEL in a subquery of the FROM clause to achieve the same result. For example, the following statement is not valid:
|
SELECT employee_id, last_name FROM employees
WHERE (employee_id, LEVEL)
IN (SELECT employee_id, 2 FROM employees)
START WITH employee_id = 2
CONNECT BY PRIOR employee_id = manager_id;
|
|
But the following statement is valid because it encapsulates the query containing the LEVEL information in the FROM clause:
|
SELECT v.employee_id, v.last_name, v.lev
FROM (
SELECT employee_id, last_name, LEVEL lev
FROM employees v
START WITH employee_id = 100
CONNECT BY PRIOR employee_id = manager_id) v
WHERE (v.employee_id, v.lev) IN (
SELECT employee_id, 2 FROM employees);
|