Return a pattern multiple times - sql

Question

I have a column called comment that can contain a certain pattern as many as 5 times and I would like to return all of them. Here is the value I have in my colum:
oijdwe wbjcwe sbr ('JOJ.TTT.ABC', test) sdfjksj dlkfjs lkf qweiuh
('JOJ.TTT.123', test) oiiqwd m lskc qu i ('JOJ.TTT.452', test) ksd
sdfskq azx ('JOJ.TTT.ACD5', test)
I would like to return the following values: JOJ.TTT.ABC, JOJ.TTT.123, JOJ.TTT.452 and JOJ.TTT.ACD5
when using the following statement
select
regexp_substr(comment,'JOJ.T{3}.{4}',1,1)
,regexp_substr(comment,'JOJ.T{3}.{4}',1,2)
,regexp_substr(comment,'JOJ.T{3}.{4}',1,3)
,regexp_substr(comment,'JOJ.T{3}.{4}',1,4)
,regexp_substr(comment,'JOJ.T{3}.{4}',1,5)
from blabla
it works fine if the pattern is JOJ.TTT.XXX but not in case there are 4 caracters after "TTT.". I know this is because of the {4} I am using. How can I get the result I want?
Thanks

Answer 1

Use [^']+ to match until the next '.
select
regexp_substr(comment_,q'!JOJ.T{3}[^']+!',1,1) a,
regexp_substr(comment_,q'!JOJ.T{3}[^']+!',1,2) b,
regexp_substr(comment_,q'!JOJ.T{3}[^']+!',1,3) c,
regexp_substr(comment_,q'!JOJ.T{3}[^']+!',1,4) d,
regexp_substr(comment_,q'!JOJ.T{3}[^']+!',1,5) e
from blabla;
I have changed comment into comment_ since comment is a reserved keyword in Oracle.

Answer 2

While the . in the regular expression works for this purpose, it will also match things you don't want, since it is a wildcard character. Use \. to match a literal .. You can also use LEVEL and CONNECT BY so you needn't write out multiple calls to REGEXP_SUBSTR() explicitly:
WITH x AS (
SELECT 'oijdwe wbjcwe sbr (''JOJ.TTT.ABC'', test) sdfjksj dlkfjs lkf qweiuh (''JOJ.TTT.123'', test) oiiqwd m lskc qu i (''JOJ.TTT.452'', test) ksd sdfskq azx (''JOJ.TTT.ACD5'', test)' AS comment1
FROM dual
)
SELECT REGEXP_SUBSTR(comment1,'JOJ\.TTT\.[A-Z0-9]+', 1, LEVEL)
FROM x
CONNECT BY REGEXP_SUBSTR(comment1,'JOJ\.TTT\.[A-Z0-9]+', 1, LEVEL) IS NOT NULL;
See SQL Fiddle here.
It gets a little kludgier if you have multiple columns and want to return them all:
WITH x AS (
SELECT 1 AS id, 'oijdwe wbjcwe sbr (''JOJ.TTT.ABC'', test) sdfjksj dlkfjs lkf qweiuh (''JOJ.TTT.123'', test) oiiqwd m lskc qu i (''JOJ.TTT.452'', test) ksd sdfskq azx (''JOJ.TTT.ACD5'', test)' AS comment1
FROM dual
UNION ALL
SELECT 2 AS id, 'oijdwe wbjcwe sbr (''JOJ.TTT.ABC'', test) sdfjksj dlkfjs lkf qweiuh (''JOJ.TTT.123'', test) oiiqwd m lskc qu i (''JOJ.TTT.452'', test) ksd sdfskq azx (''JOJ.TTT.ACD5'', test)' AS comment1
FROM dual
)
SELECT id, REGEXP_SUBSTR(comment1,'JOJ\.TTT\.[A-Z0-9]+', 1, LEVEL)
FROM x
CONNECT BY REGEXP_SUBSTR(comment1,'JOJ\.TTT\.[A-Z0-9]+', 1, LEVEL) IS NOT NULL
AND PRIOR id = id
AND PRIOR DBMS_RANDOM.VALUE IS NOT NULL;
See SQL Fiddle here. Without the PRIOR clauses Oracle will return a cross join of the values of id and of comment1.
By the way, COMMENT is an Oracle reserved word and the above queries did not work in 11g r2 with a column named comment. That is why I used comment1 instead.
Alternately, you could use [^'']+ instead of [A-Z0-9]+ but given your example data I'm not sure that's what you want.

Answer 3

I found how to fix it
select
regexp_substr(comment,'JOJ.T[^'']+',1,1)
,regexp_substr(comment,'JOJ.T[^'']+',1,2)
,regexp_substr(comment,'JOJ.T[^'']+',1,3)
,regexp_substr(comment,'JOJ.T[^'']+',1,4)
,regexp_substr(comment,'JOJ.T[^'']+',1,5)
from blabla