split out file name from path in postgres

Question

I have a field that contains windows file paths, like so:

\\\\fs1\\foo\\bar\\snafu.txt
c:\\this\\is\\why\\i\\drink\\snafu.txt
\\\\fs2\\bippity\\baz.zip
\\\\f         


        

Answer 1

CREATE OR REPLACE FUNCTION basename(text) RETURNS text
    AS $basename$
declare
    FILE_PATH alias for $1;
    ret         text;
begin
    ret := regexp_replace(FILE_PATH,'^.+[/\\]', '');
    return ret;
end;
$basename$ LANGUAGE plpgsql;

Answer 2

You can easily strip the path up to the last directory separator with an expression like

regexp_replace(path, '^.+[/\\]', '')

This will match the ocassional forward slashes produced by some software as well. Then you just count the remaining file names like

WITH files AS (
    SELECT regexp_replace(my_path, '^.+[/\\]', '') AS filename
    FROM my_table
)
SELECT filename, count(*) AS count
FROM files
GROUP BY filename
HAVING count(*) >= 2;

Answer 3

select regexp_replace(path_field, '.+/', '') from files_table;