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Oracle中检查外键是否有索引的SQL脚本分享

2024-08-29 13:58:16
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这篇文章主要介绍了Oracle中检查外键是否有索引的SQL脚本分享,本文给出了两个版本的脚本源码,一个查询所有用户,一个查询单用户,需要的朋友可以参考下
 
 

 

复制代码代码如下:

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_POSITION <= CONS.COL_CNT 
         GROUP BY I.INDEX_NAME) 
/

在上面的基础上修改了一下,可以检查所有的用户。
复制代码代码如下:

SET linesize 400;
COLUMN OWNER format a10 word_wrapped 
COLUMN COLUMNS format a30 word_wrapped 
COLUMN TABLE_NAME format a15 word_wrapped 
COLUMN CONSTRAINT_NAME format a40 word_wrapped 
SELECT OWNER,
     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.OWNER,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 DBA_CONS_COLUMNS WHERE OWNER NOT IN ('SYS','SYSTEM','SYSMAN','HR','OE','EXFSYS','DBSNMP','MDSYS','OLAPSYS','SCOTT','EXFSYS','SH','PM','CTXSYS')) A, 
               DBA_CONSTRAINTS B 
         WHERE A.CONSTRAINT_NAME = B.CONSTRAINT_NAME 
           AND B.CONSTRAINT_TYPE = 'R' 
         GROUP BY B.OWNER,B.TABLE_NAME, B.CONSTRAINT_NAME) CONS 
 WHERE COL_CNT > ALL 
 (SELECT COUNT(*) 
          FROM DBA_IND_COLUMNS I 
         WHERE I.TABLE_NAME = CONS.TABLE_NAME AND I.TABLE_OWNER=CONS.OWNER
           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) 
/
 

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