How to extract a specific string from a column

Question

I am trying to extract a specific string that starts with 'A0' and give me back the string A0 along with the rest of 9 values from a column in a table . I need to extract an auth number in a claims notes column in a table.

I am trying to use the regexp_substr function. I want to extract any string that starts with 'A0' in the clm_notes column in the table .

SELECT 
regexp_substr(CLM_NOTES, '[^A0]+', 1,1) as auth_num
FROM claims_table

this is how the column looks in the table

Clm_notes column 
New Auth from auth - A071869573

The Desired results I want is to extract the string that starts with 'A0'

A071869573

Answer 1

You will need something like the following:

SELECT 
regexp_substr(CLM_NOTES, 'A0.*$', 1,1) as auth_num
FROM claims_table

Or perhaps if all characters after A0 are digits:

SELECT 
regexp_substr(CLM_NOTES, 'A0[0-9]*$', 1,1) as auth_num
FROM claims_table

Another faster way may be the following:

SELECT 
substr(CLM_NOTES, instr(CLM_NOTES,'A0')) as auth_num
FROM claims_table

Answer 2

Rather than match what you want, match what you don't want and delete it:

select
    regexp_replace(CLM_NOTES, '.*- A0', 'A0') as auth_num
from claims_table

This matches everything before and including '- A0' and replaces it with just 'A0' , effectively deleting everything preceding 'A0' .