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本篇內容主要講解“oracle 中 SQL 全表掃描過程分析”,感興趣的朋友不妨來看看。本文介紹的方法操作簡單快捷,實用性強。下面就讓丸趣 TV 小編來帶大家學習“oracle 中 SQL 全表掃描過程分析”吧!
以下 SQL 走了全表掃描,效率下降,而 SQL 中謂詞字段選擇性非常低,通過直方圖,并從 btree 轉 bitmap 后性能提供,于是對此過程進行分析。
Select Count(*) From pmc.DesignXXXXX t Where 1=1 and OrganId= C00000220 And CategoryCode=2 and IsEnable=1 and isdelete=0 or (PublicStatus=1 and isdelete=0 );
COUNT(*)
----------
1845
較差的執行計劃:通過掃描表方式,邏輯讀需要 844525:
=====================================================
Execution Plan
----------------------------------------------------------
Plan hash value: 527126818
-----------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
-----------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1| 19 | 229K (1)| 00:45:58 |
| 1 | SORT AGGREGATE | | 1| 19 | | |
|* 2 | TABLE ACCESS FULL| DESIGNXXXXX | 4744K| 85M| 229K (1)| 00:45:58 |
-----------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
2 - filter( ISDELETE =0 AND ( PUBLICSTATUS =1 OR ORGANID = C00000220
AND CATEGORYCODE =2 AND ISENABLE =1))
Statistics
----------------------------------------------------------
1 recursive calls
0 db block gets
844525 consistent gets
842418 physical reads
0 redo size
527 bytes sent via SQL*Net to client
520 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
該 SQL 是如何選擇的執行計劃 (通過 10053 進行追蹤):
oracle 進行了次以下幾種方式的 cost 比較:
1. 評估通過全表掃描需要的 cost 是 229760.92.
Access Path: TableScan
Cost: 229760.92 Resp: 229760.92 Degree: 0
Cost_io: 229075.00 Cost_cpu: 25302994949
Resp_io: 229075.00 Resp_cpu: 25302994949
2. 評估通過位圖索引的方式 cost 是 741028, 這里是已經同時用 bitmap 方式將 or 兩邊進行聯結的消耗。
****** trying bitmap/domain indexes ******
….
Bitmap nodes:
Used IND_DESIGNXXXXX_ISENABLE_ORG
Cost = 35.099036, sel = 0.000494
Used IND_DESIGNXXXXX_CATEGORYCODE
Cost = 1281.621955, sel = 0.034894
Bitmap nodes:
Used IND_PUBLICSTATUS
Cost = 17275.447942, sel = 0.471383
Used bitmap node
Bitmap nodes:
Used bitmap node
Access path: Bitmap index – accepted
Cost: 741028.481879 Cost_io: 740534.527080 Cost_cpu: 18221443693.247154 Sel: 0.471392
因為該語句中存在 or,即分別計算 or 左右的訪問路徑消耗,再來進行組合。
3.or 右邊通過 IND_PUBLICSTATUS 索引范圍掃描 cost 是 429957
Access Path: index (AllEqRange)
Index: IND_PUBLICSTATUS
resc_io: 429587.00 resc_cpu: 13681713060
ix_sel: 0.477347 ix_sel_with_filters: 0.477347
Cost: 429957.89 Resp: 429957.89 Degree: 1
4.or 左邊分別計算使用以下索引的的消耗
1)DESIGNXXXXX_TIME_ORGANID 的消耗是 88778。
Access Path: index (SkipScan)
Index: DESIGNXXXXX_TIME_ORGANID
resc_io: 88761.00 resc_cpu: 643271006
ix_sel: 0.000509 ix_sel_with_filters: 0.000509
Cost: 88778.44 Resp: 88778.44 Degree: 1
2)IND_DESIGNXXXXX_CATEGORYCODE 的消耗是 32961.
Access Path: index (AllEqRange)
Index: IND_DESIGNXXXXX_CATEGORYCODE
resc_io: 32934.00 resc_cpu: 1020893102
ix_sel: 0.036885 ix_sel_with_filters: 0.036885
Cost: 32961.67 Resp: 32961.67 Degree: 1
ColGroup Usage:: PredCnt: 2 Matches Full: #2 Partial: Sel: 0.0005
ColGroup Usage:: PredCnt: 2 Matches Full: #2 Partial: Sel: 0.0005
3)IND_DESIGNXXXXX_CATEGORYCODE 的消耗是 32961.
Access Path: index (AllEqRange)
Index: IND_DESIGNXXXXX_ISENABLE_ORG
resc_io: 6499.00 resc_cpu: 57845156
ix_sel: 0.000494 ix_sel_with_filters: 0.000494
Cost: 6500.57 Resp: 6500.57 Degree: 1
4) 單獨 IND_DESIGNXXXXX_ISENABLE_ORG 和 IND_DESIGNXXXXX_CATEGORYCODE 轉 bitmap 的消耗是 1406。
Bitmap nodes:
Used IND_DESIGNXXXXX_ISENABLE_ORG
Cost = 35.099036, sel = 0.000494
Used IND_DESIGNXXXXX_CATEGORYCODE
Cost = 1281.621955, sel = 0.034894
Access path: Bitmap index – accepted
Cost: 1406.374238 Cost_io: 1399.626467 Cost_cpu: 248917754.369408 Sel: 0.000017
這里需要注意的是將 or 左右兩邊分別拿出來計算,最終合并需要統計計算兩邊的消耗,因此以上的所有消耗評估是:
全表掃描 (Cost: 229760.92) IND_PUBLICSTATUS 索引(Cost: 429957.89)+ 任意左邊任意一種訪問路徑方式 兩邊直接轉位圖聯結的方式(Cost: 741028)
于是自然而然選擇了全表掃描:
Final cost for query block SEL$1 (#0) – All Rows Plan:
Best join order: 1
Cost: 229760.9246 Degree: 1 Card: 1845.0000 Bytes: 35055
Resc: 229760.9246 Resc_io: 229075.0000 Resc_cpu: 25302994949
Resp: 229760.9246 Resp_io: 229075.0000 Resc_cpu: 25302994949
我們要知道以上都只是 oracle CBO 評估的結果,而在日常應用中 CBO 如果獲取的表信息不夠準確便為導致評估結果不一定是正確,而我們有時無法控制的是 SQL 每次硬解析時獲取信息是否足夠準確,這也是因此偶爾會出現執行計劃突變的狀況。
以上 SQL 通過收集直方圖后便可暫時得到解決。
這是收集直方圖后,較優的執行計劃:分別通過 btree 索引轉成 BITMAP 索引方式,邏輯讀需要 2196
================================================================
Execution Plan
----------------------------------------------------------
Plan hash value: 4067119963
--------------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 19 | 647 (1)| 00:00:08 |
| 1 | SORT AGGREGATE | | 1 | 19 | | |
|* 2 | TABLE ACCESS BY INDEX ROWID | DESIGNXXXXX | 1901 | 36119 | 647 (1)| 00:00:08 |
| 3 | BITMAP CONVERSION TO ROWIDS | | | | | |
| 4 | BITMAP OR | | | | | |
| 5 | BITMAP CONVERSION FROM ROWIDS | | | | | |
|* 6 | INDEX RANGE SCAN | IND_PUBLICSTATUS | | | 6 (0)| 00:00:01 |
| 7 | BITMAP AND | | | | | |
| 8 | BITMAP CONVERSION FROM ROWIDS| | | | | |
|* 9 | INDEX RANGE SCAN | IND_DESIGNXXXXx_ISENABLE_ORG | | | 3 (0)| 00:00:01 |
| 10 | BITMAP CONVERSION FROM ROWIDS| | | | | |
|* 11 | INDEX RANGE SCAN | IND_DESIGNXXXXXX_CATEGORYCODE | | | 102 (0)| 00:00:02 |
--------------------------------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
2 - filter(ISDELETE =0)
6 - access(PUBLICSTATUS =1)
9 - access(ISENABLE =1 AND ORGANID = C00000220)
11 - access(CATEGORYCODE =2)
Statistics
----------------------------------------------------------
1 recursive calls
0 db block gets
2196 consistent gets
0 physical reads
0 redo size
527 bytes sent via SQL*Net to client
520 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
以上 BITMAP CONVERSION 的順序過程:
步驟 1.sql 通過 IND_PUBLICSTATUS 索引到表中獲取符合條件的行,然后從獲取的行中的 rowid 轉換成 bitmap,這一步是 BITMAP CONVERSION FROM ROWIDS。
步驟 2.sql 通過 IND_DESIGNXXXXX_CATEGORYCODE 索引到表中獲取符合條件的行,然后同樣從獲取的行中的 rowid 轉換成 bitmap,這一步是 BITMAP CONVERSION FROM ROWIDS。
步驟 3.sql 通過 IND_DESIGNXXXXX_ISENABLE_ORG 索引到表中獲取符合條件的行,然后同樣從獲取的行中的 rowid 轉換成 bitmap,這一步是 BITMAP CONVERSION FROM ROWIDS。
步驟 4.sql 將步驟 2 和步驟 3 所得 bitmap 數據通過 BITMAP AND 方式取交集。
步驟 5.sql 將步驟 1 所得 bitmaps 數據與步驟 4 通過 BITMAP OR 方式取并集。
步驟 6.sql 將步驟 5 最終獲取的并集 bitmap 數據轉換成 ROWIDS,這一步是 BITMAP CONVERSION TO ROWIDS。
步驟 7.sql 將步驟 6 獲取的 rowid 通過回表方式到表中獲取所需要的字段數據,這一步是 ABLE ACCESS BY INDEX ROWID。
為什么會這樣:
當對表中的唯一度不高的列建立了 index,oracle 就有可能選擇轉為 bitmap 來執行。查看 sql 中 where 條件后字段都是選擇性非常的低。
相應字段選擇性:
COLUMN_NAME NUM_ROWS CARDINALITY SELECTIVITY
------------------------------ ---------- ----------- -----------
ORGANID 21095783 2070 .01
CATEGORYCODE 21095783 29 0
ISENABLE 21095783 2 0
ISDELETE 21095783 2 0
PUBLISHSTATE 21095783 1 0
對應索引:INDEX_NAME INDEX_COL INDEX_TYPE
-------------------------------- ---------------------- ----------------------
PMC.IND_DESIGNXXXXX_CATEGORYCODE CATEGORYCODE NORMAL-NONUNIQUE
PMC.IND_DESIGNXXXXX_ISENABLE_ORG ISENABLE,ORGANID NORMAL-NONUNIQUE
PMC.IND_PUBLICSTATUS PUBLICSTATUS NORMAL-NONUNIQUE
同樣使用 10053 追蹤增加直方圖后 SQL 執行,此時 CBO 為什么可以選擇到轉位圖的執行計劃,發現增加直方圖之后評估消耗只需要 647,而在此之前所需消耗要高達 741028。
增加直方圖后的評估:
Access path: Bitmap index – accepted
Cost: 647.047103 Cost_io: 646.348285 Cost_cpu: 25778603.541021 Sel: 0.000103
對比未增加直方圖之前的評估:
Access path: Bitmap index – accepted
Cost: 741028.481879 Cost_io: 740534.527080 Cost_cpu: 18221443693.247154 Sel: 0.471392
為什么收集直方圖后評估的消耗可以這么低?
在 oracle CBO 計算 cost 主要是 IO 成本 +CPU 成本,在計算成本之前,CBO 會收集以下統計信息:
列中不同值的數量也就是 NDV
列中的最小值 / 最大值
列中 null 值的數量
數據分布
直方圖信息(前提是收集直方圖)
對比收集直方圖前后的字段信息:
收集直方圖之前的字段信息:
Column (#4): ORGANID(
AvgLen: 10 NDV: 2023 Nulls: 4717 Density: 0.000494
Column (#29): CATEGORYCODE(
AvgLen: 2 NDV: 27 Nulls: 1164044 Density: 0.037037 Min: 0 Max: 66
Column (#38): ISENABLE(
AvgLen: 2 NDV: 2 Nulls: 1151627 Density: 0.500000 Min: 0 Max: 1
Column (#14): ISDELETE(
AvgLen: 3 NDV: 2 Nulls: 0 Density: 0.500000 Min: 0 Max: 1
Column (#32): PUBLICSTATUS(
AvgLen: 2 NDV: 2 Nulls: 1151554 Density: 0.500000 Min: 0 Max: 1
收集直方圖之后的字段信息:
Single Table Cardinality Estimation for DESIGNXXXXX[T]
Column (#14):
NewDensity:0.041803, OldDensity:0.000000 BktCnt:6033548, PopBktCnt:6033548, PopValCnt:2, NDV:2
Column (#14): ISDELETE(
AvgLen: 3 NDV: 2 Nulls: 0 Density: 0.041803 Min: 0 Max: 1
Histogram: Freq #Bkts: 2 UncompBkts: 6033548 EndPtVals: 2
Column (#4):
NewDensity:0.000185, OldDensity:0.001779 BktCnt:254, PopBktCnt:160, PopValCnt:25, NDV:2027
Column (#4): ORGANID(
AvgLen: 10 NDV: 2027 Nulls: 4830 Density: 0.000185
Histogram: HtBal #Bkts: 254 UncompBkts: 254 EndPtVals: 120
Column (#29):
NewDensity:0.000000, OldDensity:0.000000 BktCnt:5680066, PopBktCnt:5680055, PopValCnt:16, NDV:27
Column (#29): CATEGORYCODE(
AvgLen: 2 NDV: 27 Nulls: 1162620 Density: 0.000000 Min: 0 Max: 66
Histogram: Freq #Bkts: 27 UncompBkts: 5680066 EndPtVals: 27
Column (#38):
NewDensity:0.000943, OldDensity:0.000000 BktCnt:5687407, PopBktCnt:5687407, PopValCnt:2, NDV:2
Column (#38): ISENABLE(
AvgLen: 2 NDV: 2 Nulls: 1150490 Density: 0.000943 Min: 0 Max: 1
Histogram: Freq #Bkts: 2 UncompBkts: 5687407 EndPtVals: 2
ColGroup (#2, Index) IND_DESIGNXXXXX_ISENABLE_ORG
Col#: 4 38 CorStregth: 2.00
ColGroup (#3, Index) IND_DESIGNXXXXX_AUTHOR_TIME
Col#: 6 7 CorStregth: -1.00
ColGroup (#1, Index) DESIGNXXXXX_TIME_ORGANID
Col#: 4 7 CorStregth: -1.00
ColGroup Usage:: PredCnt: 3 Matches Full: Partial:
Column (#32):
NewDensity:0.000055, OldDensity:0.000000 BktCnt:5688611, PopBktCnt:5688611, PopValCnt:2, NDV:2
Column (#32): PUBLICSTATUS(
AvgLen: 2 NDV: 2 Nulls: 1150387 Density: 0.000055 Min: 0 Max: 1
Histogram: Freq #Bkts: 2 UncompBkts: 5688611 EndPtVals: 2
在沒有收集直方圖之前,發現有部分字段的 Density 都是 0.5,這個值是從 1 /NDV(基數) 得到的,這是因為 CBO 有時無法正確的統計到表的數據分布,但當收集直方圖后該值就改變了,因為在一個表中,不一定所有的數據都能分配平均,直方圖的作用就是能找出這種不平均,
那 PUBLICSTATUS 字段來說,我們看到 NDV 是 2,即是說全表之后兩個值,這兩個值是 0 或 1,在沒有收集直方圖之前 CBO 可能會認為 0 和 1 的分布是各一半,此時他去評估訪問該字段的路徑可能是全表掃描比較好,
而實際上,表中 PUBLICSTATUS=1 的數據量非常少。
sys@LVDB SQL Select Count(*) From pmc.DesignXXXXX t where PublicStatus=1 and isdelete=0 ;
COUNT(*)
———-
1845
但直到 PUBLICSTATUS 的數據分布后,CBO 評估通過 IND_PUBLICSTATUS 索引訪問 cost 只需要 6。這也是為什么收集直方圖后能更加準確的評估訪問表的消耗了。
Access Path: index (AllEqRange)
Index: IND_PUBLICSTATUS
resc_io: 6.00 resc_cpu: 457729
ix_sel: 0.000112 ix_sel_with_filters: 0.000112
Cost: 6.01 Resp: 6.01 Degree: 0
然后該種 0 或 1 的情況選擇了轉換成 bitmap 索引的模式。
其實如果不選擇 btree 轉換 bitmap 方式,直接使用 btree 索引回表效率也是沒問題的,只是需要將 sql 中的 or 拆成 union 語句
Execution Plan
----------------------------------------------------------
Plan hash value: 3766559296
------------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
------------------------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 13 | 105 (2)| 00:00:02 |
| 1 | SORT AGGREGATE | | 1 | 13 | | |
| 2 | VIEW | | 2 | 26 | 105 (2)| 00:00:02 |
| 3 | SORT UNIQUE | | 2 | 22 | 105 (2)| 00:00:02 |
| 4 | UNION-ALL | | | | | |
| 5 | SORT AGGREGATE | | 1 | 17 | 9 (12)| 00:00:01 |
|* 6 | TABLE ACCESS BY INDEX ROWID| DESIGXXXXXXX | 1 | 17 | 8 (0)| 00:00:01 |
|* 7 | INDEX RANGE SCAN | IND_DESIGNXXXXXX_ISENABLE_ORG | 6 | | 3 (0)| 00:00:01 |
| 8 | SORT AGGREGATE | | 1 | 5 | 96 (2)| 00:00:02 |
|* 9 | TABLE ACCESS BY INDEX ROWID| DESIGNXXXXXXX | 1874 | 9370 | 95 (0)| 00:00:02 |
|* 10 | INDEX RANGE SCAN | IND_PUBLICSTATUS | 2046 | | 6 (0)| 00:00:01 |
------------------------------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
6 - filter(CATEGORYCODE =2 AND ISDELETE =0)
7 - access(ISENABLE =1 AND ORGANID = C00000281)
9 - filter(ISDELETE =0)
10 - access(PUBLICSTATUS =1)
Statistics
----------------------------------------------------------
1 recursive calls
0 db block gets
2114 consistent gets
0 physical reads
0 redo size
527 bytes sent via SQL*Net to client
520 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
1 sorts (memory)
0 sorts (disk)
1 rows processed
對于開啟直方圖和 btree 轉 Bitma 都各自存在某些 bug, 有時甚至可能引發異常的性能問題,這點是需要重點注意的。
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