我已经改变了你的约束添加的“ON DELETE CASCADE”,没有它,Oracle将引发一个错误。(对于外键冲突的默认值是删除限制)
我相信回答你的问题是“否”,Oracle不会对未索引的外键列提出警告。在实践中,大多数这样的列都被编入索引,因为这是您将如何将父母加入孩子的方式。
如果你想证明给别人,不具有一定的折射率会导致锁定问题和升级(不是东西非常可取的),你可以简单地禁用表锁,并显示错误。
SQL> alter table child disable table lock;
Table altered.
SQL> delete from parent where parent_id = 10;
delete from parent where parent_id = 10
*
ERROR at line 1:
ORA-00069: cannot acquire lock -- table locks disabled for CHILD
而对于解释计划的问题,正如其他人所指出的那样,SQL从子表中删除是一个递归SQL和解释计划没有显示。
如果您跟踪会话,你会看到递归SQL。
1* alter session set SQL_TRACE = TRUE
SQL>/
Session altered.
SQL> delete from parent where parent_id = 10;
1 row deleted.
SQL> commit;
Commit complete.
SQL> alter session set SQL_TRACE=FALSe;
Session altered.
=====================
PARSING IN CURSOR #2 len=39 dep=0 uid=65 oct=7 lid=65 tim=763167901560 hv=3048246147 ad='3160891c'
delete from parent where parent_id = 10
END OF STMT
PARSE #2:c=0,e=61,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,tim=763167901555
=====================
PARSING IN CURSOR #1 len=48 dep=1 uid=0 oct=7 lid=0 tim=763167976106 hv=2120075951 ad='26722c20'
delete from "RC"."CHILD" where "PARENT_ID" = :1
END OF STMT
PARSE #1:c=0,e=42,p=0,cr=0,cu=0,mis=0,r=0,dep=1,og=4,tim=763167976100
EXEC#1:c=0,e=291,p=0,cr=7,cu=7,mis=0,r=2,dep=1,og=4,tim=763168080347
EXEC#2:c=0,e=130968,p=0,cr=8,cu=14,mis=0,r=1,dep=0,og=1,tim=763168091605
STAT #2 id=1 cnt=1 pid=0 pos=1 obj=0 op='DELETE PARENT (cr=8 pr=0 pw=0 time=130887 us)'
STAT #2 id=2 cnt=1 pid=1 pos=1 obj=58703 op='INDEX UNIQUE SCAN SYS_C006951 (cr=1 pr=0 pw=0 time=19 us)'
STAT #1 id=1 cnt=0 pid=0 pos=1 obj=0 op='DELETE CHILD (cr=7 pr=0 pw=0 time=233 us)'
STAT #1 id=2 cnt=2 pid=1 pos=1 obj=58704 op='TABLE ACCESS FULL CHILD (cr=7 pr=0 pw=0 time=76 us)'
相关链接:http://www.oracle-base.com/articles/10g/SQLTrace10046TrcsessAndTkprof10g.php
我不知道如果你有机会到这个与否,但甲骨文的SQL分析(http://docs.oracle.com/html/A86647_01/vmqintro.htm )具有一些用于SQL分析和调优的强大功能。 – 2012-02-15 21:41:56
@ZackMacomber谢谢,我不熟悉它,但我会确保我学习。它实际上会做我所问的吗? – 2012-02-15 22:36:30