How to find the starting position of the last word from the string in PostgreSQL

Question

What will be the Postgres equivalent for the below code from Oracle.

Select instr('abc de fgh',' ', -1) from dual;

returns: 7

Original code: substr(country, instr(country,' ', -1)+1);

I want to create the same logic in Postgres but position & reverse function didn't work

Answer 1

You apparently want the last element from the string where elements are delimited by a space character.

If you are using Postgres 14, you can use:

split_part('abc de fgh', ' ', -1);

which returns fgh

Or with a column reference: split_part(country, ' ', -1)

In earlier versions, split_part() doesn't allow negative offsets, so you would need something different:

(string_to_array('abc de fgh', ' '))[cardinality(string_to_array('abc de fgh', ' '))] 

It's a bit shorter with a column reference:

(string_to_array(country, ' '))[cardinality(string_to_array(country, ' '))] 

This first converts the string into an array, then picks the last element of the array (which in turn is determined using the cardinality() function that returns the length of the array)

Answer 2

Since you mention REVERSE, you can use:

SELECT RIGHT(value, POSITION(' ' in REVERSE(value)) - 1) AS country,
       LENGTH(value) - POSITION(' ' in REVERSE(value)) + 1 AS pos
FROM   (SELECT 'abc de fgh' AS value) v;

Which outputs:

country	pos
fgh	7

db<>fiddle here