sql split string by space into table in postgresql

Question

I looking for a function like regexp_split_to_table, but our db is version 8.2.9, so it doesn't have it. I'm really only splitting on a space, so a string like

how now brown cow

would return

+------+
|Column|
+------+
|how   | 
|now   | 
|brown | 
|cow   |
+------+

is there a simple function that can handle this, or something I have to write myself?

Answer 1

You can split an array to a resultset by using the unnest function, and you can turn a string literal into an array by using the string_to_array function. Combine both and you get this:

alvherre=# select unnest(string_to_array('the quick lazy fox', ' '));
 unnest 
--------
 the
 quick
 lazy
 fox
(4 filas)

Since 8.2 does not have UNNEST, you can write it in PostgreSQL like this:

create or replace function unnest(anyarray) returns setof anyelement
language sql as $$
   select $1[i] from generate_series(array_lower($1, 1),
                                     array_upper($1, 1)) as i;
$$; 

Answer 2

I think you'll have to RETURNS SET or RETURNS TABLE yourself.

Updated answer: using PL/pgSQL:

pg=> CREATE OR REPLACE FUNCTION string_to_rows(text) RETURNS SETOF TEXT AS $$ 
  DECLARE
    elems text[];      
  BEGIN
    elems := string_to_array($1, ' ');
    FOR i IN array_lower(elems, 1) .. array_upper(elems, 1) LOOP
      RETURN NEXT elems[i];
    END LOOP;
    RETURN;
  END
$$ LANGUAGE 'plpgsql';
CREATE FUNCTION

pg=> SELECT "Column" FROM string_to_rows('how now brown cow') d("Column");
 Column 
--------
 how
 now
 brown
 cow
(4 rows)

Original answer: using PL/perl:

pg=> CREATE LANGUAGE plperl; 
CREATE LANGUAGE

pg=> CREATE FUNCTION psplit_to_rows(text) RETURNS SETOF TEXT AS $$
pg$>   for my $t (split ' ', $_[0]) { return_next $t; }
pg$>   undef;
pg$> $$ LANGUAGE plperl;
CREATE FUNCTION

pg=> SELECT "Column" FROM psplit_to_rows('how now brown cow') d("Column");
 Column 
--------
 how
 now
 brown
 cow
(4 rows)

Obviously you can extend this to handle a delimiter of your choosing, etc. (Note, I'm not sure if you really wanted that column named "Column", requiring identifier quoting to avoid keyword clash, but, there you are.)