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Oracle® Database SQL Language Reference
11g Release 1 (11.1)

Part Number B28286-01
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Hierarchical Queries

If a table contains hierarchical data, then you can select rows in a hierarchical order using the hierarchical query clause:

hierarchical_query_clause::=

Description of hierarchical_query_clause.gif follows
Description of the illustration hierarchical_query_clause.gif

START WITH specifies the root row(s) of the hierarchy.

CONNECT BY specifies the relationship between parent rows and child rows of the hierarchy.

Both the CONNECT BY condition and the PRIOR expression can take the form of an uncorrelated subquery. However, CURRVAL and NEXTVAL are not valid PRIOR expressions, so the PRIOR expression cannot refer to a sequence.

You can further refine a hierarchical query by using the CONNECT_BY_ROOT operator to qualify a column in the select list. This operator extends the functionality of the CONNECT BY [PRIOR] condition of hierarchical queries by returning not only the immediate parent row but all ancestor rows in the hierarchy.

See Also:

CONNECT_BY_ROOT for more information about this operator and "Hierarchical Query Examples"

Oracle processes hierarchical queries as follows:

Oracle then uses the information from these evaluations to form the hierarchy using the following steps:

  1. Oracle selects the root row(s) of the hierarchy--those rows that satisfy the START WITH condition.

  2. Oracle selects the child rows of each root row. Each child row must satisfy the condition of the CONNECT BY condition with respect to one of the root rows.

  3. Oracle selects successive generations of child rows. Oracle first selects the children of the rows returned in step 2, and then the children of those children, and so on. Oracle always selects children by evaluating the CONNECT BY condition with respect to a current parent row.

  4. If the query contains a WHERE clause without a join, then Oracle eliminates all rows from the hierarchy that do not satisfy the condition of the WHERE clause. Oracle evaluates this condition for each row individually, rather than removing all the children of a row that does not satisfy the condition.

  5. Oracle returns the rows in the order shown in Figure 9-1. In the diagram, children appear below their parents. For an explanation of hierarchical trees, see Figure 3-1, "Hierarchical Tree".

Figure 9-1 Hierarchical Queries

Description of Figure 9-1 follows
Description of "Figure 9-1 Hierarchical Queries"

To find the children of a parent row, Oracle evaluates the PRIOR expression of the CONNECT BY condition for the parent row and the other expression for each row in the table. Rows for which the condition is true are the children of the parent. The CONNECT BY condition can contain other conditions to further filter the rows selected by the query. The CONNECT BY condition cannot contain a subquery.

If the CONNECT BY condition results in a loop in the hierarchy, then Oracle returns an error. A loop occurs if one row is both the parent (or grandparent or direct ancestor) and a child (or a grandchild or a direct descendent) of another row.

Note:

In a hierarchical query, do not specify either ORDER BY or GROUP BY, as they will override the hierarchical order of the CONNECT BY results. If you want to order rows of siblings of the same parent, then use the ORDER SIBLINGS BY clause. See order_by_clause .

Hierarchical Query Examples

CONNECT BY Example The following hierarchical query uses the CONNECT BY clause to define the relationship between employees and managers:

SELECT employee_id, last_name, manager_id
   FROM employees
   CONNECT BY PRIOR employee_id = manager_id
   ORDER BY employee_id, last_name, manager_id;

EMPLOYEE_ID LAST_NAME                 MANAGER_ID
----------- ------------------------- ----------
        100 King
        101 Kochhar                          100
        101 Kochhar                          100
        102 De Haan                          100
        102 De Haan                          100
        103 Hunold                           102
        103 Hunold                           102
        103 Hunold                           102
        104 Ernst                            103
        104 Ernst                            103
        104 Ernst                            103
...

LEVEL Example The next example is similar to the preceding example, but uses the LEVEL pseudocolumn to show parent and child rows:

SELECT employee_id, last_name, manager_id, LEVEL
   FROM employees
   CONNECT BY PRIOR employee_id = manager_id
   ORDER BY employee_id, last_name, manager_id, LEVEL;

EMPLOYEE_ID LAST_NAME                 MANAGER_ID      LEVEL
----------- ------------------------- ---------- ----------
        100 King                                          1
        101 Kochhar                          100          1
        101 Kochhar                          100          2
        102 De Haan                          100          1
        102 De Haan                          100          2
        103 Hunold                           102          1
        103 Hunold                           102          2
        103 Hunold                           102          3
        104 Ernst                            103          1
        104 Ernst                            103          2
        104 Ernst                            103          3
...

START WITH Examples The next example adds a START WITH clause to specify a root row for the hierarchy and an ORDER BY clause using the SIBLINGS keyword to preserve ordering within the hierarchy:

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;

LAST_NAME                 EMPLOYEE_ID MANAGER_ID      LEVEL
------------------------- ----------- ---------- ----------
King                              100                     1
Cambrault                         148        100          2
Bates                             172        148          3
Bloom                             169        148          3
Fox                               170        148          3
Kumar                             173        148          3
Ozer                              168        148          3
Smith                             171        148          3
De Haan                           102        100          2
Hunold                            103        102          3
Austin                            105        103          4
Ernst                             104        103          4
Lorentz                           107        103          4
Pataballa                         106        103          4
Errazuriz                         147        100          2
Ande                              166        147          3
Banda                             167        147          3
...

In the hr.employees table, the employee Steven King is the head of the company and has no manager. Among his employees is John Russell, who is the manager of department 80. If you update the employees table to set Russell as King's manager, you create a loop in the data:

UPDATE employees SET manager_id = 145
   WHERE employee_id = 100;

SELECT last_name "Employee", 
   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;

ERROR:
ORA-01436: CONNECT BY loop in user data

The NOCYCLE parameter in the CONNECT BY condition causes Oracle to return the rows in spite of the loop. The CONNECT_BY_ISCYCLE pseudocolumn shows you which rows contain the cycle:

SELECT last_name "Employee", 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
   ORDER BY "Employee", "Cycle", LEVEL, "Path";

Employee                       Cycle      LEVEL Path
------------------------- ---------- ---------- -------------------------
Abel                               0          3 /King/Zlotkey/Abel
Ande                               0          3 /King/Errazuriz/Ande
Banda                              0          3 /King/Errazuriz/Banda
Bates                              0          3 /King/Cambrault/Bates
Bernstein                          0          3 /King/Russell/Bernstein
Bloom                              0          3 /King/Cambrault/Bloom
Cambrault                          0          2 /King/Cambrault
Cambrault                          0          3 /King/Russell/Cambrault
Doran                              0          3 /King/Partners/Doran
Errazuriz                          0          2 /King/Errazuriz
Fox                                0          3 /King/Cambrault/Fox
...

CONNECT_BY_ROOT Examples 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:

SELECT last_name "Employee", CONNECT_BY_ROOT last_name "Manager",
   LEVEL-1 "Pathlen", SYS_CONNECT_BY_PATH(last_name, '/') "Path"
   FROM employees
   WHERE LEVEL > 1 and department_id = 110
   CONNECT BY PRIOR employee_id = manager_id
   ORDER BY "Employee", "Manager", "Pathlen", "Path";

Employee        Manager            Pathlen Path
--------------- --------------- ---------- ------------------------------
Gietz           Higgins                  1 /Higgins/Gietz
Gietz           King                     3 /King/Kochhar/Higgins/Gietz
Gietz           Kochhar                  2 /Kochhar/Higgins/Gietz
Higgins         King                     2 /King/Kochhar/Higgins
Higgins         Kochhar                  1 /Kochhar/Higgins

The following example uses a GROUP BY clause to return the total salary of each employee in department 110 and all employees below that employee in the hierarchy:

SELECT name, SUM(salary) "Total_Salary" FROM (
   SELECT CONNECT_BY_ROOT last_name as name, Salary
      FROM employees
      WHERE department_id = 110
      CONNECT BY PRIOR employee_id = manager_id)
      GROUP BY name
   ORDER BY name, "Total_Salary";

NAME                      Total_Salary
------------------------- ------------
Gietz                             8300
Higgins                          20300
King                             20300
Kochhar                          20300

See Also: