将查询Oracle转换为Postgresql

Question

In Oracle

(select regexp_substr('a,,b','[^,]+', 1, level) 
 from dual 
 connect by regexp_substr('a,,b', '[^,]+', 1, level) is not null);

it gives me output

a
b

In Postgresql

select regexp_split_to_table( 'a,,b',',');

it gives me output

a

b

It gives a blank row between a and b . Can anyone please suggest how I can get output like Oracle .

Answer 1

In oracle you were searching for matches, not splitting. Doing the same in Postgres:

SELECT m[1] FROM regexp_matches('a,,b', '[^,]+', 'g') AS T(m);

EDIT: Thanks to a_horse_with_no_name and TimBiegeleisen for making me notice it was returning an array in each row.

Answer 2

For a simple splitting/unnesting like that, string_to_array would be a better choice. Processing regular expressions is quite expensive.

If you put the set returning function into the FROM clause (where it should be) you can filter out the empty rows:

SELECT t.m
FROM unnest(string_to_array('a,,b', ',')) as t(m)
where nullif(trim(t.m),'') is not null;

nullif(trim(tm),'') will also handle inputs like a, ,b and treat the empty string as null.

Online example: https://rextester.com/VDWUEM85289