No show string between round brackets SQL Oracle

Question

I have a table with these rows as example:

City:

 San Francisco (CA) Miami (FL) 

As a result I want this:

City:

 San Francisco Miami 

Any help?

Regards.

Answer 1

You can use REGEX_REPLACE :

CREATE TABLE tab(City VARCHAR(120));

INSERT INTO tab
VALUES ('San Francisco (CA)');

INSERT INTO tab
VALUES ('Miami (FL)');

SELECT TRIM(REGEXP_REPLACE(city, '\((.+?)\)', '')) AS City
FROM tab;

SqlFiddleDemo

If you want to UPDATE you can use:

UPDATE tab
SET City = TRIM(REGEXP_REPLACE(city, '\((.+?)\)', ''))
WHERE INSTR(City, '(') > 0;

SELECT City
FROM tab;

SqlFiddleDemo2

Answer 2

For the sake of argument, here's a REGEXP_SUBSTR( ) example. It assumes you want to keep everything from the start of the strin up to but not including the space before the first open paren.

SQL> with tbl(city) as (
     select 'San Francisco (CA)' from dual
     union
     select 'Miami (FL)' from dual
   )
   select regexp_substr(city, '^(.*) \(.*$', 1, 1, null, 1) new_city
   from tbl;

NEW_CITY
------------------
Miami
San Francisco

SQL>

EDIT: Added a REGEXP_REPLACE solution that does not require a TRIM():

with tbl(city) as (
  select 'San Francisco (CA)' from dual
  union
  select 'Miami (FL)' from dual
)
select regexp_replace(city, '^(.*) \(.*$', '\1') new_city
from tbl;

Answer 3

Or you could do something like

SELECT CASE
         WHEN INSTR(CITY, '(') > 2 THEN
           SUBSTR(CITY, 1, INSTR(CITY, '(')-2)
         ELSE CITY
       END AS CITY
  FROM CITIES;

SQLFiddle here

Best of luck.