편집 : 보너스 쿼리 @Rob van Wijk에서 (재귀) CTE, inspired by a note을 사용하여 11g을 필요로 (R2?) :
SQL> WITH data AS (
2 SELECT ans_code, Ans_Desc, tag_name, tag_value,
3 row_number() OVER (partition BY ans_code ORDER BY t.rowid) no,
4 row_number() OVER
5 (partition BY ans_code ORDER BY t.rowid DESC) is_last
6 FROM answer_step_dtl a
7 JOIN tag_mst t ON a.ans_desc LIKE '%' || t.tag_name || '%'
8 ), n(ans_code, no, is_last, replaced) AS (
9 SELECT ans_code, no n, is_last,
10 replace (ans_desc, tag_name, tag_value) replaced
11 FROM data
12 WHERE no = 1
13 UNION ALL
14 SELECT d.ans_code, d.no, d.is_last,
15 replace (n.replaced, d.tag_name, d.tag_value) replaced
16 FROM data d
17 JOIN n ON d.ans_code = n.ans_code
18 AND d.no = n.no + 1
19 )
20 SELECT *
21 FROM n
22 WHERE is_last=1;
ANS_CODE NO IS_LAST REPLACED
-------------------- -- ------- ---------------------------------------
40000000000000000164 1 1 Enter <B>EXAMPLE</B> in connection
50000000000000005770 1 1 Enter <B>EXAMPLE.COM</B> and press Ok.
40000000000000000165 2 1 Enter <B>EXAMPLE</B> in EXAMPLE.COM.
초기 대답 :
당신은 PL/SQL 함수를 사용할 수 있습니다. 다음과 같이 여러 개의 태그를 대체하더라도 작동합니다 :
CREATE OR REPLACE FUNCTION replacetags(p_desc VARCHAR2)
RETURN VARCHAR2 IS
l_result LONG := p_desc;
l_tag_pos INTEGER := 1;
l_tag tag_mst.tag_name%TYPE;
BEGIN
LOOP
l_tag := regexp_substr(l_result, '<[^<]+>', l_tag_pos);
l_tag_pos := regexp_instr(l_result, '<[^<]+>', l_tag_pos) + 1;
EXIT WHEN l_tag IS NULL;
BEGIN
SELECT replace(l_result, l_tag, tag_value)
INTO l_result
FROM tag_mst
WHERE tag_name = l_tag;
EXCEPTION
WHEN no_data_found THEN
NULL; -- tag doesn't exist in tag_mst
END;
END LOOP;
RETURN l_result;
END;
SQL> SELECT ans_code, replacetags(ans_desc)
2 FROM answer_step_dtl;
ANS_CODE REPLACETAGS(ANS_DESC)
--------------------- ----------------------------------------
50000000000000005770 Enter <B>EXAMPLE.COM</B> and press Ok.
40000000000000000164 Enter <B>EXAMPLE</B> in connection
40000000000000000165 Enter <B>EXAMPLE</B> in EXAMPLE.COM.
동일한 'ans_desc'에 두 개의 태그 이름이있을 수 있습니까? – Ben
예, 가능하면 @joe 솔루션이 완벽합니다. –
@ 벤 @ 내 질문에 친절하게 그것을 보았습니다. –