我正在尝试减少SQL使用的缓冲区大小。 以下是解释计划。
看一下解释计划,我倾向于关注下面的第3行
3 NESTED LOOPS Cost: 1,407 Bytes: 11,902,228 Cardinality: 276,796
因为成本从55(第2行)跳到1407
2 INDEX RANGE SCAN INDEX (UNIQUE) SIEBEL.S_PROD_LN_PROD_U1 Cost: 55 Bytes: 247,359 Cardinality: 11,779
我对此几乎没有问题:
提前多多感谢。
Plan
SELECT STATEMENT HINT: FIRST_ROWSCost: 2,101 Bytes: 2,563 Cardinality: 1
32 NESTED LOOPS Cost: 2,101 Bytes: 2,563 Cardinality: 1
29 NESTED LOOPS Cost: 2,097 Bytes: 2,518 Cardinality: 1
27 NESTED LOOPS Cost: 2,095 Bytes: 2,500 Cardinality: 1
24 HASH JOIN RIGHT ANTI Cost: 2,092 Bytes: 2,444 Cardinality: 1
4 VIEW VIEW SYS.VW_NSO_1 Cost: 1,407 Bytes: 8,857,472 Cardinality: 276,796
3 NESTED LOOPS Cost: 1,407 Bytes: 11,902,228 Cardinality: 276,796
1 TABLE ACCESS FULL TABLE SIEBEL.S_PROD_LN Cost: 83 Bytes: 528 Cardinality: 24
2 INDEX RANGE SCAN INDEX (UNIQUE) SIEBEL.S_PROD_LN_PROD_U1 Cost: 55 Bytes: 247,359 Cardinality: 11,779
23 NESTED LOOPS OUTER Cost: 104 Bytes: 39,200 Cardinality: 28
20 NESTED LOOPS OUTER Cost: 50 Bytes: 37,604 Cardinality: 28
17 NESTED LOOPS OUTER Cost: 50 Bytes: 36,540 Cardinality: 28
14 NESTED LOOPS OUTER Cost: 23 Bytes: 35,700 Cardinality: 28
11 NESTED LOOPS OUTER Cost: 23 Bytes: 34,300 Cardinality: 28
8 HASH JOIN RIGHT OUTER Cost: 23 Bytes: 32,900 Cardinality: 28
5 TABLE ACCESS FULL TABLE SIEBEL.S_PROD_INT_CRSE Cost: 2 Bytes: 32,718 Cardinality: 82
7 TABLE ACCESS BY INDEX ROWID TABLE SIEBEL.S_PROD_INT Cost: 20 Bytes: 11,252 Cardinality: 29
6 INDEX RANGE SCAN INDEX SIEBEL.S_PROD_INT_X2_X Cost: 3
10 TABLE ACCESS BY INDEX ROWID TABLE SIEBEL.S_LIT Cost: 0 Bytes: 50 Cardinality: 1
9 INDEX UNIQUE SCAN INDEX (UNIQUE) SIEBEL.S_LIT_P1 Cost: 0 Cardinality: 1
13 TABLE ACCESS BY INDEX ROWID TABLE SIEBEL.S_LIT Cost: 0 Bytes: 50 Cardinality: 1
12 INDEX UNIQUE SCAN INDEX (UNIQUE) SIEBEL.S_LIT_P1 Cost: 0 Cardinality: 1
16 TABLE ACCESS BY INDEX ROWID TABLE SIEBEL.S_PROD_LN Cost: 1 Bytes: 30 Cardinality: 1
15 INDEX UNIQUE SCAN INDEX (UNIQUE) SIEBEL.S_PROD_LN_P1 Cost: 0 Cardinality: 1
19 TABLE ACCESS BY INDEX ROWID TABLE SIEBEL.S_PROD_LN Cost: 0 Bytes: 38 Cardinality: 1
18 INDEX UNIQUE SCAN INDEX (UNIQUE) SIEBEL.S_PROD_LN_P1 Cost: 0 Cardinality: 1
22 TABLE ACCESS BY INDEX ROWID TABLE SIEBEL.S_PROD_INT Cost: 2 Bytes: 57 Cardinality: 1
21 INDEX UNIQUE SCAN INDEX (UNIQUE) SIEBEL.S_PROD_INT_P1 Cost: 1 Cardinality: 1
26 TABLE ACCESS BY INDEX ROWID TABLE SIEBEL.S_VOD Cost: 3 Bytes: 56 Cardinality: 1
25 INDEX RANGE SCAN INDEX (UNIQUE) SIEBEL.S_VOD_U2 Cost: 2 Cardinality: 1
28 INDEX RANGE SCAN INDEX SIEBEL.S_ISS_OBJ_DEF_M3 Cost: 2 Bytes: 36 Cardinality: 2
31 TABLE ACCESS BY INDEX ROWID TABLE SIEBEL.S_VOD_VER Cost: 4 Bytes: 45 Cardinality: 1
30 INDEX RANGE SCAN INDEX (UNIQUE) SIEBEL.S_VOD_VER_U1 Cost: 2 Cardinality: 1
答案 0 :(得分:0)
可以在此处找到有关如何解释Oracle解释计划的详细说明:http://www.akadia.com/services/ora_interpreting_explain_plan.html
在不了解底层数据库结构,表卷或查询这些表的SQL语句的情况下解释您的解释计划将非常困难。