Unindexed foreign keys

Unindexed foreign keys makes the child table get locked and also it makes the performance down.
There is a very beautiful article written by Tom kyte on Unindexed foreign kyes.
The link is here

and if you are not able to access, the content goes here
-------------------------------------------------------------------------------
Unindexed Foreign Keys

Having Unindexed foreign keys can be a performance issue. There are two issues associated with unindexed foreign keys. The first is the fact that a table lock will result if you update the parent records primary key (very very unusual) or if you delete the parent record and the child's foreign key is not indexed.

To read about this issue, please see the Concepts Guide the section on Maintaining Data Integrity/Concurrency Control, Indexes, and Foreign Keys.

The second issue has to do with performance in general of a parent child relationship. Consider that if you have an on delete cascade and have not indexed the child table (eg: EMP is child of DEPT. Delete deptno = 10 should cascade to EMP. If deptno in emp is not indexed -- full table scan). This full scan is probably undesirable and if you delete many rows from the parent table, the child table will be scanned once for each parent row deleted.

Also consider that for most (not all, most) parent child relationships, we query the objects from the 'master' table to the 'detail' table. The glaring exception to this is a code table (short code to long description). For master/detail relationships, if you do not index the foreign key, a full scan of the child table will result.

So, how do you easily discover if you have unindexed foreign keys in your schema? This script can help. When you run it, it will generate a report such as:


SQL> @unindex

STAT TABLE_NAME COLUMNS COLUMNS
---- ------------------------------ -------------------- --------------------
**** APPLICATION_INSTANCES AI_APP_CODE
ok EMP DEPTNO DEPTNO



The **** in the first row shows me that I have an unindexed foreign key in the table APPLICATION_INSTANCES. The ok in the second row shows me I have a table EMP with an indexed foreign key.
------------------------------------------------------------------------

The script that Tom is used can be found here to find the unindexed foreign keys

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)


Thanks Tom for all the help