Extract number from string with Oracle function

You'd use REGEXP_REPLACE in order to remove all non-digit characters from a string:

select regexp_replace(column_name, '[^0-9]', '')
from mytable;

or

select regexp_replace(column_name, '[^[:digit:]]', '')
from mytable;

Of course you can write a function extract_number. It seems a bit like overkill though, to write a funtion that consists of only one function call itself.

create function extract_number(in_number varchar2) return varchar2 is
begin
  return regexp_replace(in_number, '[^[:digit:]]', '');
end; 

You can use regular expressions for extracting the number from string. Lets check it. Suppose this is the string mixing text and numbers 'stack12345overflow569'. This one should work:

select regexp_replace('stack12345overflow569', '[[:alpha:]]|_') as numbers from dual;

which will return "12345569".

also you can use this one:

select regexp_replace('stack12345overflow569', '[^0-9]', '') as numbers,
       regexp_replace('Stack12345OverFlow569', '[^a-z and ^A-Z]', '') as characters
from dual

which will return "12345569" for numbers and "StackOverFlow" for characters.

This works for me, I only need first numbers in string:

TO_NUMBER(regexp_substr(h.HIST_OBSE, '\.*[[:digit:]]+\.*[[:digit:]]*'))

the field had the following string: "(43 Paginas) REGLAS DE PARTICIPACION".

result field: 43