楼上的试验很能说明问题, 应用正好有这方面的问题, 学习了.
Tom的一段sql, 作用是找出没有建立索引的外键, 太复杂, 我是没有看懂
[PHP]COLUMN COLUMNS format a30 word_wrapped
COLUMN tablename format a15 word_wrapped
COLUMN constraint_name format a15 word_wrapped
SELECT TABLE_NAME,
CONSTRAINT_NAME,
CNAME1 || NVL2(CNAME2, ',' || CNAME2, NULL) ||
NVL2(CNAME3, ',' || CNAME3, NULL) ||
NVL2(CNAME4, ',' || CNAME4, NULL) ||
NVL2(CNAME5, ',' || CNAME5, NULL) ||
NVL2(CNAME6, ',' || CNAME6, NULL) ||
NVL2(CNAME7, ',' || CNAME7, NULL) ||
NVL2(CNAME8, ',' || CNAME8, NULL) COLUMNS
FROM (SELECT B.TABLE_NAME,
B.CONSTRAINT_NAME,
MAX(DECODE(POSITION, 1, COLUMN_NAME, NULL)) CNAME1,
MAX(DECODE(POSITION, 2, COLUMN_NAME, NULL)) CNAME2,
MAX(DECODE(POSITION, 3, COLUMN_NAME, NULL)) CNAME3,
MAX(DECODE(POSITION, 4, COLUMN_NAME, NULL)) CNAME4,
MAX(DECODE(POSITION, 5, COLUMN_NAME, NULL)) CNAME5,
MAX(DECODE(POSITION, 6, COLUMN_NAME, NULL)) CNAME6,
MAX(DECODE(POSITION, 7, COLUMN_NAME, NULL)) CNAME7,
MAX(DECODE(POSITION, 8, COLUMN_NAME, NULL)) CNAME8,
COUNT(*) COL_CNT
FROM (SELECT SUBSTR(TABLE_NAME, 1, 30) TABLE_NAME,
SUBSTR(CONSTRAINT_NAME, 1, 30) CONSTRAINT_NAME,
SUBSTR(COLUMN_NAME, 1, 30) COLUMN_NAME,
POSITION
FROM USER_CONS_COLUMNS) A,
USER_CONSTRAINTS B
WHERE A.CONSTRAINT_NAME = B.CONSTRAINT_NAME
AND B.CONSTRAINT_TYPE = 'R'
GROUP BY B.TABLE_NAME, B.CONSTRAINT_NAME) CONS
WHERE COL_CNT > ALL
(SELECT COUNT(*)
FROM USER_IND_COLUMNS I
WHERE I.TABLE_NAME = CONS.TABLE_NAME
AND I.COLUMN_NAME IN (CNAME1, CNAME2, CNAME3, CNAME4, CNAME5,
CNAME6, CNAME7, CNAME8)
AND I.COLUMN_POSITIONALL
(SELECT COUNT(*)
FROM USER_IND_COLUMNS I
WHERE I.TABLE_NAME = CONS.TABLE_NAME
AND I.COLUMN_NAME IN (CNAME1, CNAME2, CNAME3, CNAME4, CNAME5,
CNAME6, CNAME7, CNAME8)
AND I.COLUMN_POSITION <= CONS.COL_CNT
GROUP BY I.INDEX_NAME)
/[/PHP] |
