Retrieving not full result / regexp substr

Question

Hi ,
when I try to run this code
SQL> select
2 regexp_substr('But , soft! What light through yonder window breaks?',
3 'l[[:alpha:]] {4}') AS result
4 from dual;
I get just this ( R
where the result should be like the full sub string which is ( light) not just R.
Ml

Answer 1

Mr.lonely wrote:
Hi ,
when I try to run this code
SQL> select
2 regexp_substr('But , soft! What light through yonder window breaks?',
3 'l[[:alpha:]] {4}') AS resultThere's a space right before '{4}', so you're looking for a lower-case 'l', followed by any letter, followed by exactly 4 spaces.
4 from dual;
I get just this ( R
where the result should be like the full sub string which is ( light) not just R. Actually, you're not even getting 'R' in the result set. (REGEXP_SUBSTR can't return anything that's not in the 1st argument, and there's no capital 'R' in that argument.) The "R" you see is the first character of the column header, RESULT. REGEXP_SUBSTR is returning NULL, because the pattern (ending in 4 spaces) wasn't found.
MlLose the space:
SELECT  REGEXP_SUBSTR ( 'But , soft! What light through yonder window breaks?'
                , 'l[[:alpha:]]{4}'
                )            AS result
FROM    dual
;or, if you like having the space there, use the optional 5th argument to REGEXP_SUBSTR to ignore that space:
SELECT  REGEXP_SUBSTR ( 'But , soft! What light through yonder window breaks?'
                , 'l[[:alpha:]] {4}'
                , 1
                , 1
                , 'x'     -- x means ignore whitespace in 2nd argument
                )            AS result
FROM    dual
;In order to pass the 5th argument in SQL, you have to pass the 3rd and 4th arguments, too.