找出哪些表中未加索引的外键

下面的例子展示了如何使用这个脚本来找出表C中未加索引的外键：

ops$tkyte@ORA10G> column columns format a30 word_wrapped
ops$tkyte@ORA10G> column tablename format a15 word_wrapped
ops$tkyte@ORA10G> column constraint_name format a15 word_wrapped
ops$tkyte@ORA10G> select table_name, constraint_name,
2 cname1 || nvl2(cname2,','||cname2,null) ||
3 nvl2(cname3,','||cname3,null) || nvl2(cname4,','||cname4,null) ||
4 nvl2(cname5,','||cname5,null) || nvl2(cname6,','||cname6,null) ||
5 nvl2(cname7,','||cname7,null) || nvl2(cname8,','||cname8,null)
6 columns
7 from ( select b.table_name,
8 b.constraint_name,
9 max(decode( position, 1, column_name, null )) cname1,
10 max(decode( position, 2, column_name, null )) cname2,
264 / 860
11 max(decode( position, 3, column_name, null )) cname3,
12 max(decode( position, 4, column_name, null )) cname4,
13 max(decode( position, 5, column_name, null )) cname5,
14 max(decode( position, 6, column_name, null )) cname6,
15 max(decode( position, 7, column_name, null )) cname7,
16 max(decode( position, 8, column_name, null )) cname8,
17 count(*) col_cnt
18 from (select substr(table_name,1,30) table_name,
19 substr(constraint_name,1,30) constraint_name,
20 substr(column_name,1,30) column_name,
21 position
22 from user_cons_columns ) a,
23 user_constraints b
24 where a.constraint_name = b.constraint_name
25 and b.constraint_type = 'R'
26 group by b.table_name, b.constraint_name
27 ) cons
28 where col_cnt > ALL
29 ( select count(*)
30 from user_ind_columns i
31 where i.table_name = cons.table_name
32 and i.column_name in (cname1, cname2, cname3, cname4,
33 cname5, cname6, cname7, cname8 )
265 / 860
34 and i.column_position <= cons.col_cnt
35 group by i.index_name
36 )
37 /
TABLE_NAME CONSTRAINT_NAME COLUMNS
------------------------------ --------------- ------------------------------
C SYS_C009485 X

所以，这个脚本展示出，表C在列X上有一个外键，但是没有索引。通过对X加索引，就可以完全消除这个锁定问题。