Get a word after specific word using regexp_substr in sql oracle

Question

I have tried

select regexp_substr (\'sys: error: This is a message \'\'123:\'\' for column EMP_NB\', \'[[:alpha:]_]+\',1,9) from dual

I have c

Answer 1

You may use

select regexp_substr ('sys: error: This is a message ''123:'' for column EMP_NB', 'column[[:space:]]*([[:alpha:]_]+)', 1, 1, NULL, 1) from dual

Here,

• column - matches column word
• [[:space:]]* - 0 or more whitespace chars
• ([[:alpha:]_]+) - captures into Group 1 any one or more letters or underscores.

The value captured is returned only, since the last group ID argument is set to 1.

Answer 2

Just anchor the pattern to the end of the string:

select regexp_substr('sys: error: This is a message ''123:'' for column EMP_NB', '[[:alpha:]_]+$')
from dual

Here is a db<>fiddle.

Your example data has a period at the end. If that is part of the string, then you can use regexp_replace():

select regexp_replace(message, '^.*[^[:alpha:]_]([[:alpha:]_]+)[^[:alpha:]_]*$', '\1')
from (select 'sys: error: This is a message ''123:'' for column EMP_NB' as message from dual union all
      select 'sys: error: This is a message ''45346:'' for column EM_NM.' as message from dual union all
      select 'sys: error: This is a message ''78324f9:'' for column DEPT_NO_VL.' as message from dual
     ) x