Order Oracle Database Results as Parent And then Its Children, Then Order the Children By Date

Question

Consider the following table:

employer_id	supervisor_employer_ref	date_hired
22436	NULL	01/02/2006
37854	25558	07/12/2018
25558	NULL	10/16/2010
48796	25558	11/11/2015
28449	22436	08/14/2016
34663	25558	10/10/2019
45687	22436	04/12/2020

For some specific reason i have to order this table by the employer and then its supervised employers and order them by date_hired on an Oracle 12c DB. I'm aware this is not the most ideal way of making a database but for a specific reason I have to keep it this way.

Intended result:

employer_id	supervisor_employer_ref	date_hired
22436	NULL	01/02/2006
28449	22436	08/14/2016
45687	22436	04/12/2020
25558	NULL	10/16/2010
48796	25558	11/11/2015
37854	25558	07/12/2018
34663	25558	10/10/2019
45687	22436	04/12/2020

Answer 1

You want a hierarchical query and to use ORDER SIBLINGS BY :

SELECT *
FROM   table_name
START WITH
       supervisor_employer_ref IS NULL
CONNECT BY
       PRIOR employer_id = supervisor_employer_ref
ORDER SIBLINGS BY date_hired;

Which, for the sample data:

CREATE TABLE table_name (employer_id, supervisor_employer_ref, date_hired) AS
SELECT 22436, NULL,  TO_DATE( '01/02/2006', 'MM/DD/YYYY' ) FROM DUAL UNION ALL
SELECT 37854, 25558, TO_DATE( '07/12/2018', 'MM/DD/YYYY' ) FROM DUAL UNION ALL
SELECT 25558, NULL,  TO_DATE( '10/16/2010', 'MM/DD/YYYY' ) FROM DUAL UNION ALL
SELECT 48796, 25558, TO_DATE( '11/11/2015', 'MM/DD/YYYY' ) FROM DUAL UNION ALL
SELECT 28449, 22436, TO_DATE( '08/14/2016', 'MM/DD/YYYY' ) FROM DUAL UNION ALL
SELECT 34663, 25558, TO_DATE( '10/10/2019', 'MM/DD/YYYY' ) FROM DUAL UNION ALL
SELECT 45687, 22436, TO_DATE( '04/12/2020', 'MM/DD/YYYY' ) FROM DUAL;

Outputs:

EMPLOYER_ID	SUPERVISOR_EMPLOYER_REF	DATE_HIRED
22436		02-JAN-06
28449	22436	14-AUG-16
45687	22436	12-APR-20
25558		16-OCT-10
48796	25558	11-NOV-15
37854	25558	12-JUL-18
34663	25558	10-OCT-19

db<>fiddle here