Oracle中利用10053事件来分析Oracle是如何做出最终的执行计划(三)

2014-11-24 18:33:17 · 作者: · 浏览: 17
orted: cost > best plan cost
***********************
(newjo-stop-1) k:0, spcnt:0, perm:2, maxperm:2000
*********************************
Number of join permutations tried: 2
*********************************
(newjo-save) [1 0 ]
Final - All Rows Plan: Best join order: 1
Cost: 5.5199 Degree: 1 Card: 94.0000 Bytes: 940
Resc: 5.5199 Resc_io: 5.0000 Resc_cpu: 3025807
Resp: 5.5199 Resp_io: 5.0000 Resc_cpu: 3025807
kkoipt: Query block SEL$1 (#0)
******* UNPARSED QUERY IS *******
SELECT "T2"."RN" "RN","T2"."NAME" "NAME" FROM "YORKER"."T1" "T1","YORKER"."T2" "T2" WHERE "T1"."RN"<100 AND "T2"."RN"<100 AND "T1"."RN"="T2"."RN"
kkoqbc-end
: call(in-use=31732, alloc=0), compile(in-use=33436, alloc=0)
apadrv-end: call(in-use=31732, alloc=0), compile(in-use=34024, alloc=0)

sql_id=ar0vn3xs804bs.
Current SQL statement for this session:
explain plan for select t2.* from t1,t2 where t1.rn <100 and t1.rn=t2.rn

============
Plan Table
============
------------------------------------------------+-----------------------------------+
| Id | Operation | Name | Rows | Bytes | Cost | Time |
------------------------------------------------+-----------------------------------+
| 0 | SELECT STATEMENT | | | | 6 | |
| 1 | HASH JOIN | | 94 | 940 | 6 | 00:00:01 |
| 2 | INDEX RANGE SCAN | IND_T1 | 95 | 380 | 2 | 00:00:01 |
| 3 | TABLE ACCESS BY INDEX ROWID | T2 | 99 | 594 | 3 | 00:00:01 |
| 4 | INDEX RANGE SCAN | IND_T2 | 99 | | 2 | 00:00:01 |
------------------------------------------------+-----------------------------------+