Get second match from regexp_matches results - sql

Question

I have a name column which looks like this:
'1234567 - 7654321 - some - more - text'
I need to get a string "7654321". I am stuck with the following:
SELECT regexp_matches('1234567 - 7654321 - some - more - text', '\d+', 'g');
regexp_matches
----------------
{1234567}
{7654321}
(2 rows)
How do I what I want? Maybe there's a better option than regexp_matches - gladly will consider. Thx!

Answer 1

You could use REGEXP_REPLACE:
SELECT REGEXP_REPLACE('1234567 - 7654321 - some - more - text', '^\d+[^\d]+(\d+).*$', '\1');
Output
7654321
This regexp looks for a string starting with some number of digits (^\d+) followed by some non-digit characters ([^\d]+) and then another group of digits ((\d+)) followed by some number of characters until the end of the string (.*$). The () around the second group of digit characters makes that a capturing group, which we can then refer to in the replacement string with \1. Since REGEXP_REPLACE only replaces the parts of the string that match the regex, it is necessary to have a regex that matches the whole string in order to replace it with just the desired data.
Update
If there are potentially characters before the first set of digits, you should change the regex to
^[^\d]*\d+[^\d]+(\d+).*$
Update 2
If it's possible that there is only one set of numbers at the beginning, we must make matching the first part optional. We can do that with a non-capturing group:
^[^\d]*(?:\d+[^\d]+)?(\d+).*$
This makes the match on the first set of digits optional so that if it doesn't exist (i.e. there is only one set of digits) the regex will still match. By using a non-capturing group (adding the ?: to the beginning of the group, we don't need to change the replacement string from \1. Updated SQLFiddle

Answer 2

regexp_matches() returns a table, so you can use that in the from clause together with the with ordinality option:
SELECT t.value
from regexp_matches('1234567 - 7654321 - some - more - text', '\d+', 'g') with ordinality as t(value,idx)
where t.idx = 2;
Note that value is still an array, to get the actual array element you can use:
SELECT t.value[1]
from regexp_matches('1234567 - 7654321 - some - more - text', '\d+', 'g') with ordinality as t(value,idx)
where t.idx = 2;