Oracle SQL REGEX extracting text from the middle of a varchar2

Question

I'm looking to extract a bit of text from the middle of a varchar2 column.

Here are a few examples:

 TEST DATA - SCOTLAND 1A
 TEST DATA - ENGLAND 6A
 TEST DATA - WALES 3A
 TEST DATA - IRELAND 2A

The data I'm looking to return would be:-

 SCOTLAND
 ENGLAND
 WALES
 IRELAND

Many thanks

Lee

Answer 1

This query seems to be working:

SELECT
    input,
    REGEXP_REPLACE(input, '.*- (\D+).*', '\1') AS country
FROM yourTable;

Demo

Here we use REGEXP_REPLACE with the pattern:

.*- (\D+).*

This captures any non digit characters occurring between the em dash and the space that follows the country name. Then, we just replace with that captured country name.

Answer 2

This could be a way, assuming that you need to get the part of the string between the (unique) '-' and the first digit.

with testTable(string) as (
    select 'TEST DATA - SCOTLAND 1A' from dual union all
    select 'TEST DATA - ENGLAND 6A' from dual union all
    select 'TEST DATA - WALES 3A' from dual union all
    select 'TEST DATA - IRELAND 2A' from dual union all
    select 'TEST DATA - NORTHERN IRELAND 2A' from dual
)
select string,
       regexp_substr(string, '\- ([^0-9]*)[0-9]', 1, 1, 'i', 1) result
from testTable



STRING                          RESULT                         
------------------------------- -------------------------------
TEST DATA - SCOTLAND 1A         SCOTLAND                       
TEST DATA - ENGLAND 6A          ENGLAND                        
TEST DATA - WALES 3A            WALES                          
TEST DATA - IRELAND 2A          IRELAND                        
TEST DATA - NORTHERN IRELAND 2A NORTHERN IRELAND               

5 rows selected.

Answer 3

Thanks to Aleksej for CTE.

How about good, old SUBSTR + INSTR ?

SQL> with testTable(string) as (
  2      select 'TEST DATA - SCOTLAND 1A' from dual union all
  3      select 'TEST DATA - ENGLAND 6A' from dual union all
  4      select 'TEST DATA - WALES 3A' from dual union all
  5      select 'TEST DATA - IRELAND 2A' from dual union all
  6      select 'TEST DATA - NORTHERN IRELAND 2A' from dual
  7  )
  8  select
  9    string,
 10    trim(substr(string,
 11                instr(string, '-') + 1,
 12                instr(string, ' ', -1) - instr(string, '-')
 13         )) result
 14  from testtable;

STRING                          RESULT
------------------------------- -------------------------------
TEST DATA - SCOTLAND 1A         SCOTLAND
TEST DATA - ENGLAND 6A          ENGLAND
TEST DATA - WALES 3A            WALES
TEST DATA - IRELAND 2A          IRELAND
TEST DATA - NORTHERN IRELAND 2A NORTHERN IRELAND

SQL>

Answer 4

One option would be applying firstly regexp_replace() as pattern, which looks for digits, to see the country names as the third words and the regexp_substr() to extract those ones only.

with cte as
(
 select 'TEST DATA - SCOTLAND 1A' as str from dual union all 
 select 'TEST DATA - ENGLAND 6A'         from dual union all 
 select 'TEST DATA - WALES 3A'           from dual union all 
 select 'TEST DATA - IRELAND 2A'         from dual
)    
select regexp_substr(  
                     regexp_replace(str, '-(.*)\d.*', '\1' ) 
                     , '[^ ]+', 1 , 3 )
       as "Country",
       trim(regexp_substr( str, '[^-]+(\s)', 1 , 2 ))
       as "Country2"
  from cte;

Country    Country2
--------   --------
SCOTLAND   SCOTLAND
ENGLAND    ENGLAND
WALES      WALES
IRELAND    IRELAND

Demo

PS the "column2" may be a smarter alternative without need of regexp_replace() function.