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怎樣淺析Oracle的物理讀 邏輯讀 一致性讀 當前模式讀，針對這個問題，這篇文章詳細介紹了相對應的分析和解答，希望可以幫助更多想解決這個問題的小伙伴找到更簡單易行的方法。
在ORACLE數據庫中有物理讀（Physical Reads）、邏輯讀（Logical Reads）、一致性讀（Consistant Get）、當前模式讀（DB Block Gets）等諸多概念，如果不理解或混淆這些概念的話，對你深入理解一些知識無疑是一個障礙，但是這些概念確實挺讓讓人犯暈的。下面我們總結、學習一下這方面的知識點。捋一捋他們的關系和特點，希望對你有所幫助。
物理讀（Physical Reads）
從磁盤讀取數據塊到內存的操作叫物理讀，當SGA里的高速緩存（Cache Buffer）里面不存在這些數據塊時，就會產生物理讀，另外。像全表掃描、磁盤排序等操作也可能產生物理讀，原因也是因為ORACLE數據庫需要訪問的數據塊較多，而有些數據塊不在內存當中，需要從磁盤讀取。
邏輯讀（Logical Reads）
概念1：邏輯讀指ORACLE從內存讀到的數據塊數量。一般來說， logical reads = db block gets + consistent gets
概念2：邏輯讀指的就是從Buffer Cache中讀取數據塊。按照訪問數據塊的模式不同，可以分為當前模式讀（Current Read）和一致性讀（Consistent Read）。
這兩個概念本質是一樣的，只是措辭不一樣。
一致性讀（Consistant Get）
ORACLE是一個多用戶系統。當一個會話開始讀取數據還未結束讀取之前，可能會有其他會話修改了它將要讀取的數據。如果會話讀取到修改后的數據，就會造成數據的不一致。一致性讀就是為了保證數據的一致性。在Buffer Cache中的數據塊上都會有最后一次修改數據塊時的SCN。如果一個事務需要修改數據塊中數據，會先在回滾段中保存一份修改前數據和SCN的數據塊，然后再更新Buffer Cache中的數據塊的數據及其SCN，并標識其為“臟”數據。當其他進程讀取數據塊時，會先比較數據塊上的SCN和進程自己的SCN。如果數據塊上的SCN小于等于進程本身的SCN，則直接讀取數據塊上的數據；如果數據塊上的SCN大于進程本身的SCN，則會從回滾段中找出修改前的數據塊讀取數據。通常，普通查詢都是一致性讀。
當前模式讀（DB Block Gets）
個人覺得當前模式讀（db block gets）是最難理解的一個概念，通常情況下db block gets 可以理解為是DML操作才會產生的.
當前模式讀（db block gets）即讀取數據塊是當前的最新數據。任何時候在Buffer Cache中都只有一份當前數據塊。當前讀通常發生在對數據進行修改、刪除操作時。這時，進程會給數據加上行級鎖，并且標識數據為“臟”數據。current mode產生db block gets，一般在DML操作時產生，query mode產生consistent gets（一致性讀），一般在查詢時產生。他們兩個總和一般稱為邏輯讀，logical read。
有個有意思的現象，在ask tom或一些資料中，你會發現Oracle 8i在SELECT查詢當中還能看到db block gets，但是ORACLE 10以及以上版本在SELECT語句中db block gets一般為0。
了解完了概念，如果你還是有一些疑問和不解，那我們結合實際例子來理解一下這些概念吧。如下所示:
SQL> show user;
USER is "SYS"
SQL> create table test
2 as
3 select * from dba_objects;
Table created.
SQL> alter session set sql_trace=true;
System altered.
SQL> set autotrace on;
SQL> select object_type, count(1) from test
2 group by object_type;
OBJECT_TYPE COUNT(1)
------------------- ----------
EDITION 1
INDEX PARTITION 264
CONSUMER GROUP 25
SEQUENCE 223
TABLE PARTITION 240
SCHEDULE 3
QUEUE 35
RULE 1
JAVA DATA 328
...............................
...............................
43 rows selected.
Execution Plan
----------------------------------------------------------
Plan hash value: 1435881708
---------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
---------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 75101 | 806K| 284 (2)| 00:00:04 |
| 1 | HASH GROUP BY | | 75101 | 806K| 284 (2)| 00:00:04 |
| 2 | TABLE ACCESS FULL| TEST | 75101 | 806K| 281 (1)| 00:00:04 |
---------------------------------------------------------------------------
Note
-----
- dynamic sampling used for this statement (level=2)
Statistics
----------------------------------------------------------
48 recursive calls
0 db block gets
1109 consistent gets
1029 physical reads
0 redo size
1694 bytes sent via SQL*Net to client
545 bytes received via SQL*Net from client
4 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
43 rows processed
SQL> select object_type, count(1) from test
2 group by object_type;
OBJECT_TYPE COUNT(1)
------------------- ----------
EDITION 1
INDEX PARTITION 264
CONSUMER GROUP 25
SEQUENCE 223
TABLE PARTITION 240
..............................
..............................
43 rows selected.
Execution Plan
----------------------------------------------------------
Plan hash value: 1435881708
---------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
---------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 75101 | 806K| 284 (2)| 00:00:04 |
| 1 | HASH GROUP BY | | 75101 | 806K| 284 (2)| 00:00:04 |
| 2 | TABLE ACCESS FULL| TEST | 75101 | 806K| 281 (1)| 00:00:04 |
---------------------------------------------------------------------------
Note
-----
- dynamic sampling used for this statement (level=2)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
1034 consistent gets
0 physical reads
0 redo size
1694 bytes sent via SQL*Net to client
545 bytes received via SQL*Net from client
4 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
43 rows processed
SQL> set autotrace off
SQL> alter session set sql_trace =false;
Session altered.
SQL> SELECT T.value
2 || '/'
3 || Lower(Rtrim(I.INSTANCE, Chr(0)))
4 || '_ora_'
5 || P.spid
6 || '.trc' TRACE_FILE_NAME
7 FROM (SELECT P.spid
8 FROM v$mystat M,
9 v$session S,
10 v$process P
11 WHERE M.statistic# = 1
12 AND S.sid = M.sid
13 AND P.addr = S.paddr) P,
14 (SELECT T.INSTANCE
15 FROM v$thread T,
16 v$parameter V
17 WHERE V.name = 'thread'
18 AND ( V.value = 0
19 OR T.thread# = To_number(V.value) )) I,
20 (SELECT value
21 FROM v$parameter
22 WHERE name = 'user_dump_dest') T;
TRACE_FILE_NAME
--------------------------------------------------------------------------------
/u01/app/oracle/diag/rdbms/gsp/gsp/trace/gsp_ora_24900.trc
SQL語句第一次執行時，一致性讀（consistent gets）為1109， 物理讀（physical reads）為1029，當前模式讀（db block gets）為0. 如果你再執行一次上面SQL語句，你會發現物理讀（physical reads）會降低為0了，因為上一次查詢，ORACLE已經將表test的所有數據塊讀取到buffer cache里面了。當然生產環境實際情況會復雜很多。
我們先用tkprof工具格式化一下trace文件，然后我們分析一下 out_24900.prf文件。
[oracle@DB-Server trace]$ tkprof gsp_ora_24900.trc out_24900.prf aggregate=no;
TKPROF: Release 11.2.0.1.0 - Development on Thu Sep 22 10:12:15 2016
Copyright (c) 1982, 2009, Oracle and/or its affiliates. All rights reserved.
在分析之前，我們先了解一下一些概念、術語
count = number of times OCI procedure was executed
cpu = cpu time in seconds executing
elapsed = elapsed time in seconds executing
disk = number of physical reads of buffers from disk # 物理讀
query = number of buffers gotten for consistent read # 一致性讀
current = number of buffers gotten in current mode (usually for update) # 當前模式讀
rows = number of rows processed by the fetch or execute call
call：每次SQL語句的處理都分成三個部分
Parse：這步包括語法檢查和語義檢查（包括檢查是否有正確的授權和所需要用到的表、列以及其他引用到的對象是否存在）、以及將SQL語句轉換、生成執行計劃等。
Execute：這步是真正的由ORACLE來執行語句。對于insert、update、delete操作，這步會修改數據，對于select操作，這步就只是確定選擇的記錄。
Fetch：返回查詢語句中所獲得的記錄，這步只有select語句會被執行。
count : 這個語句被parse、execute、fetch的次數。
cpu ：這個語句對于所有的parse、execute、fetch所消耗的cpu的時間，以秒為單位。
elapsed ：這個語句所有消耗在parse、execute、fetch的總的時間。
disk ：從磁盤上的數據文件中物理讀取的數據塊的數量。
query ：在一致性讀模式下，一致性讀的數量。
current ：在current模式下，即當前模式讀下db blocks gets的數量。
rows : 所有SQL語句返回的記錄數目，但是不包括子查詢中返回的記錄數目。對于select語句，返回記錄是在fetch這步，對于insert、update、delete操作，返回記錄則是在execute這步。
如下截圖所示（圖1與圖2本是連接在一起的，由于太長，分開截圖，兩張圖片有相同部分），由于我們實驗過程中，并沒有采集統計信息，你會看到trac文件里面有一個動態采樣（如果你在創建表，做一次統計信息收集，結果會有一些差別），另外，物理讀和一致性讀如下，跟上面的執行計劃中的數據一致。
disk(物理讀) = 747+282 = 1029
query(一致性讀) = 1035+74 = 1109
繼續分析格式化的prf文件，我們會看到第二次查詢的query(一致性讀)為1034， disk(物理讀)為0上面例子，讓我們了解了物理讀、一致性讀，那么接下來看看當前模式讀（db block gets）的例子
SQL> create table t
2 ( id number(10)
3 );
Table created.
SQL> set autotrace on;
SQL> insert into t
2 values(1000);
1 row created.
Execution Plan
----------------------------------------------------------
---------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
---------------------------------------------------------------------------------
| 0 | INSERT STATEMENT | | 1 | 100 | 1 (0)| 00:00:01 |
| 1 | LOAD TABLE CONVENTIONAL | T | | | | |
---------------------------------------------------------------------------------
Statistics
----------------------------------------------------------
1 recursive calls
7 db block gets
1 consistent gets
0 physical reads
748 redo size
836 bytes sent via SQL*Net to client
783 bytes received via SQL*Net from client
3 SQL*Net roundtrips to/from client
1 sorts (memory)
0 sorts (disk)
1 rows processed
SQL> insert into t
2 values(1001);
1 row created.
Execution Plan
----------------------------------------------------------
---------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
---------------------------------------------------------------------------------
| 0 | INSERT STATEMENT | | 1 | 100 | 1 (0)| 00:00:01 |
| 1 | LOAD TABLE CONVENTIONAL | T | | | | |
---------------------------------------------------------------------------------
Statistics
----------------------------------------------------------
1 recursive calls
1 db block gets
1 consistent gets
0 physical reads
308 redo size
837 bytes sent via SQL*Net to client
783 bytes received via SQL*Net from client
3 SQL*Net roundtrips to/from client
1 sorts (memory)
0 sorts (disk)
1 rows processed
一致性讀如何計算呢？
關于一致性讀如何計算呢？ 我查了一下資料，一般一致性讀consistent gets ~= numrows/arraysize + blocks ，確切的說是consistent reads計算 ~=ceil(獲取行數(card)/arraysize)+used blocks， 而且這個不是絕對等于，而是約等于的關系。 但是這個不是官方資料，而是asktom和一些技術博客的介紹，我們來驗證看看吧
SQL> exec dbms_stats.gather_table_stats(user, 'TEST');
PL/SQL procedure successfully completed.
SQL>
SQL> set autotrace traceonly stat
SQL> select * from test;
72271 rows selected.
Statistics
----------------------------------------------------------
448 recursive calls
0 db block gets
5846 consistent gets
1031 physical reads
0 redo size
8296071 bytes sent via SQL*Net to client
53521 bytes received via SQL*Net from client
4820 SQL*Net roundtrips to/from client
3 sorts (memory)
0 sorts (disk)
72271 rows processed
SQL> /
72271 rows selected.
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
5789 consistent gets
0 physical reads
0 redo size
8296071 bytes sent via SQL*Net to client
53521 bytes received via SQL*Net from client
4820 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
72271 rows processed
SQL> set autotrace off;
SQL> set serveroutput on;
SQL> exec show_space('TEST',USER);Free Blocks............................. 0
Total Blocks............................ 1,152
Total Bytes............................. 9,437,184
Total MBytes............................ 9
Unused Blocks........................... 121
Unused Bytes............................ 991,232
Last Used Ext FileId.................... 1
Last Used Ext BlockId................... 89,344
Last Used Block......................... 7
PL/SQL procedure successfully completed.
SQL> show arraysize ;
arraysize 15
SQL> select ceil(72271/15) + 1152-121 from dual;
CEIL(72271/15)+1152-121
-----------------------
5850
SQL> SELECT COUNT(DISTINCT dbms_rowid.rowid_block_number(ROWID)) AS blocks FROM TEST;
BLOCKS
----------
1030
SQL> SELECT ceil(72271/15) + 1030 FROM DUAL;
CEIL(72271/15)+1030
-------------------
5849
不管是5849還是5850，都和5879 或5846有一點的出入？也就是說上面那個公式不能用等于號，關于這個，其實在https://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:880343948514 里面，你會看到這里介紹的也是一個約等于關系，而不是一個絕對等于的關系。在這里我想深入一點，無奈知識有限。 從上面的公式， 我們可以看到一致性讀跟arraysize的關系是蠻大的。那么我們來測試驗證一下，先將araraysize調整為50
SQL> set autotrace off;
SQL> set arraysize 50
SQL> set autotrace traceonly stat;
SQL> select * from test;
72271 rows selected.
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
2456 consistent gets
0 physical reads
0 redo size
7668743 bytes sent via SQL*Net to client
16418 bytes received via SQL*Net from client
1447 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
72271 rows processed
SQL>
SQL> SELECT ceil(72271/50) + 1030 FROM DUAL;
CEIL(72271/50)+1030
-------------------
2476
SQL>
如上所示，一致性讀從5789降為了2456，有興趣的可以做一下實驗。另外，由于在Oracle中，取數據最后都是從buffer cache中取，所以每出現一個physical reads必然會出現一次 logical reads. 也就是說物理讀（physical reads）一定小于邏輯讀（logical reads=db block gets + consistent gets），也就是說物理讀一定小于一致性讀，但是也有物理讀大于邏輯讀的情況。
PURPOSE
In some circumstances, you can find that tkprof report shows more physical reads than logical reads, which isn't the current result as the physical reads are normally included in logical reads.
SCOPE & APPLICATION
This article will be useful for the DBA's and customers who are concerned by the tuning of Requests.
Why Physical reads are greater than Logical reads
Sometimes, you can find the following content in tkprof report:
· Physical Reads = Disk (total)
· Logical Reads = Query (total) + Current (total)
call | count | cpu | elapsed | disk | query | current | rows |
------- | ------ | -------- | ---------- | ---------- | ---------- | ---------- | ---------- |
Parse | 1 | 0.67 | 1.10 | 0 | 0 | 0 | 0 |
Execute | 1 | 0.00 | 0.00 | 0 | 0 | 0 | 0 |
Fetch | 2202 | 167.48 | 678.70 | 579441 | 283473 | 17418 | 33014 |
------- | ------ | -------- | ---------- | ---------- | ---------- | ---------- | ---------- |
total | 2204 | 168.15 | 679.81 | 579441 | 283473 | 17418 | 33014 |
The 'disk' column is then greater than the 'query' + 'current' columns. This isn't usual.
To find the root cause of the problem, you must generate a 10046 event trace file level 8 and check for direct read waits in it.
In 10046 raw trace, you will find "direct path read" and "direct path write" waits like the example below:
WAIT #1: nam='direct path read' ela= 10076 p1=4 p2=29035 p3=1
with P1 = file#, P2 = start block#, P3 = num blocks
The "direct path read" waits account explains the difference between logical and physical reads.
In Oracle 9.2 and above, TKProf will print waits associated with each SQL statement in the output file.
Explanation:
The reason for more physical reads than logical reads is due to the number of direct reads block access. Direct path reads are generally used by Oracle when reading directly into PGA memory (as opposed to into the buffer cache).
They may happen on different actions:
· Sort IO on disk.
· Read by parallel Query Slaves when scanning is done.
· Blocks read by anticipation (readahead).
Such reads are done without loading blocks into the Buffer Cache. They can be single or multiblock reads.
Utilizing Direct Path Reads in this manner prevents the Oracle Buffer cache from beeing overloaded.
Oracle uses this optimisation when it considers that its not necessary to share the blocks between different sessions.
最后我們來看一個，熱表上一致性讀飆漲的案例，其實這個是Oracle 9i&10g編程藝術:深入數據庫體系結構這本書籍里面的一個案例，我們在此重演一遍，希望能加深大家對一致性讀的理解，首先準備測試數據環境
SQL> show user;
USER is "TEST"
SQL> create table t( x int);
Table created.
SQL> insert into t values(1);
1 row created.
SQL> commit;
Commit complete.
SQL> exec dbms_stats.gather_table_stats(user, 'T');
PL/SQL procedure successfully completed.
SQL> set autotrace on statistics;
SQL> select * from t;
X
----------
1
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
7 consistent gets
0 physical reads
0 redo size
523 bytes sent via SQL*Net to client
523 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
SQL>
如上所示，一般情況下一致性讀為7，但是我們在一個會話窗口準備執行下面SQL，頻繁修改表T
SQL> begin
2 for i in 1 .. 100000
3 loop
4 update t set x=x+1;
5 commit;
6 end loop;
7 end;
8 /
PL/SQL procedure successfully completed.
在上面會話執行的同時，我們在另外一個會話窗口馬上執行下面SQL，你會看到一致性讀飆漲。
SQL> alter session set isolation_level=serializable;
Session altered.
SQL> set autotrace on statistics;
SQL> select * from t;
X
----------
1
Statistics
----------------------------------------------------------
1 recursive calls
0 db block gets
23681 consistent gets
0 physical reads
0 redo size
523 bytes sent via SQL*Net to client
523 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
將會話設置為使用SERIALIZABLE 隔離級別，這樣無論在會話中運行多少次查詢，都將得到事務開始時刻的查詢結果。
Oracle 9i&10g編程藝術:深入數據庫體系結構
在sql調優的時候，一個關鍵指標就是consistent gets，如果這個指標很低，一般認為sql語句執行還是很高效的，反之效率會很低。但是這個指標我們知之甚少，對于這個指標的計算方式我們也是懵懵懂懂。對于邏輯讀來說，一般都是基于Logical Reads= Consistent Gets + DB Block Gets
如果我們知道logical reads是1000，我們可能錯誤地認為查詢讀取了1000*8k(約為8M)
看了博客https://viveklsharma.wordpress.com/2010/03/04/consistent-gets-myth/后，發現自己的認識是錯誤的，也按捺不住在本地測試了一把，受益匪淺。
首先我們來創建一個表，數據量為2000條。
n1@TEST11G> create table test_consistent_get as select * from all_objects where rownum between 1 and 2000;
Table created.
然后收集統計信息
n1@TEST11G> exec dbms_stats.gather_table_stats(user,'TEST_CONSISTENT_GET');
PL/SQL procedure successfully completed.
查看相應的數據塊為30個
n1@TEST11G> select num_rows,blocks,table_name,last_analyzed,global_stats from user_tables where table_name='TEST_CONSISTENT_GET';
NUM_ROWS BLOCKS TABLE_NAME LAST_ANAL GLO
---------- ---------- ------------------------------ --------- ---
2000 30 TEST_CONSISTENT_GET 20-APR-15 YES
n1@TEST11G> set autot trace
我們來看看執行計劃，很明顯走了一個全表掃描。但是我們需要關注的是統計信息中的consistent gets
n1@TEST11G> select * from test_consistent_get;
Execution Plan
----------------------------------------------------------
Plan hash value: 1444268095
-----------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
-----------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 2000 | 164K| 10 (0)| 00:00:01 |
| 1 | TABLE ACCESS FULL| TEST_CONSISTENT_GET | 2000 | 164K| 10 (0)| 00:00:01 |
-----------------------------------------------------------------------------------------
Statistics
----------------------------------------------------------
1 recursive calls
0 db block gets
163 consistent gets
0 physical reads
0 redo size
199754 bytes sent via SQL*Net to client
1883 bytes received via SQL*Net from client
135 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
2000 rows processed
可以看到這個表占用的數據塊為30，但是consistent gets卻為163，很顯然不是說這個全表掃描向cache里讀入了163*8k的數據
我們可以通過rowid來得到對應的數據塊和其中的數據情況
n1@TEST11G> select dbms_rowid.ROWID_BLOCK_NUMBER(rowid) blkno, count(*) cnt
from test_consistent_get
group by dbms_rowid.ROWID_BLOCK_NUMBER(rowid) order by 1;
BLKNO CNT
---------- ----------
263827 88
263828 84
263829 81
263830 76
263831 81
263832 80
263833 82
263834 77
263835 73
263836 78
263837 79
263838 79
263839 81
263841 82
263842 77
263843 81
263844 80
263845 81
263846 78
263847 78
263848 76
263849 78
263850 78
263851 76
263852 81
263853 15
26 rows selected.
可以通過rowid得到相關的數據塊為26個。查看段頭，發現對應的數據塊是263826是不在上面的rowid對應的數據塊范圍內的。
n1@TEST11G> select header_block,blocks ,extents from dba_segments where segment_name='TEST_CONSISTENT_GET';
HEADER_BLOCK BLOCKS EXTENTS
------------ ---------- ----------
263826 32 4
對應的區和數據塊信息如下：
n1@TEST11G> select EXTENT_ID, FILE_ID, BLOCK_ID, BLOCKS from dba_extents where SEGMENT_NAME='TEST_CONSISTENT_GET';
EXTENT_ID FILE_ID BLOCK_ID BLOCKS
---------- ---------- ---------- ----------
0 4 263824 8
1 4 263832 8
2 4 263840 8
3 4 263848 8
下面的語句可以算出對于每個數據塊對應的consistent gets的值。
n1@TEST11G>
variable b1 number;
exec :b1:=15;
compute sum of total_cnt on report
compute sum of touch_cnt on report
break on report
select blkno, total_cnt, final_cnt, rows_remaining,
case when rows_remaining=0 then touch_cnt+1 else touch_cnt end touch_cnt
from (
select blkno, total_cnt, final_cnt, rows_remaining,
case when total_cnt = final_cnt then ceil(final_cnt/:b1) else ceil(final_cnt/:b1)+1 end touch_cnt
from (
select blkno, cnt total_cnt,
case when rownum=1 or lag(rows_remaining) over (order by blkno)=0
then cnt else (cnt-(:b1-lag(rows_remaining) over (order by blkno))) end final_cnt,
rows_remaining
from (
select blkno, cnt, rr,
lead(rr) over(order by blkno) next_rr,
lead(blkno) over(order by blkno) next_blk,
ceil(rr/:b1) touch_cnt,
mod(rr,:b1) rows_remaining
from (
select dbms_rowid.ROWID_BLOCK_NUMBER(rowid) blkno, count(*) cnt,
sum(count(*)) over(order by dbms_rowid.ROWID_BLOCK_NUMBER(rowid)) rr
from test_consistent_get
group by dbms_rowid.ROWID_BLOCK_NUMBER(rowid) order by 1))));
BLKNO TOTAL_CNT FINAL_CNT ROWS_REMAINING TOUCH_CNT
---------- ---------- ---------- -------------- ----------
263827 88 88 13 6
263828 84 82 7 7
263829 81 73 13 6
263830 76 74 14 6
263831 81 80 5 7
263832 80 70 10 6
263833 82 77 2 7
263834 77 64 4 6
263835 73 62 2 6
263836 78 65 5 6
263837 79 69 9 6
263838 79 73 13 6
263839 81 79 4 7
263841 82 71 11 6
263842 77 73 13 6
263843 81 79 4 7
263844 80 69 9 6
263845 81 75 0 7
263846 78 78 3 6
263847 78 66 6 6
263848 76 67 7 6
263849 78 70 10 6
263850 78 73 13 6
263851 76 74 14 6
263852 81 80 5 7
263853 15 5 5 2
---------- ----------
sum 2000 159
可以看到對于這個全表掃描的場景，consistent gets不是衡量對于cache的數據塊數而是次數。
比如對于上面的數據塊263827 ，數據條數為88條，arraysize為15，則可以簡單說明一下是如何計算這個consistent gets值的。
對于數據塊263827,放入PGA中，得到了15行，這個時候可以理解為consistent gets=1
對于數據塊263827,再次從PGA中得到，得到了15行，這個時候consistent gets=2
依次類推
對于數據塊263827,再次從PGA中得到，得到了13行，這個時候consistent gets=6
或者也可以基本按照這個公式來計算,數據行數/arraysize+數據塊數=consistent gets
比如這個例子，2000/15+30 大概是163.3左右，所以163還是靠譜的。
對于arraysize未20,30,的情況下，相應的consistent gets也會減少。簡單模擬一下。
n1@TEST11G> set arraysize 20
n1@TEST11G> set autot trace exp stat
n1@TEST11G> select *from test_consistent_get;
Statistics
----------------------------------------------------------
1 recursive calls
0 db block gets
128 consistent gets
0 physical reads
0 redo size
195334 bytes sent via SQL*Net to client
1509 bytes received via SQL*Net from client
101 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
2000 rows processed
n1@TEST11G> set autot off
n1@TEST11G> select 2000/20+30 from dual;
2000/20+30
----------
130
n1@TEST11G> set arraysize 30
n1@TEST11G> set autot trace stat
n1@TEST11G> select *from test_consistent_get;
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
96 consistent gets
0 physical reads
0 redo size
191044 bytes sent via SQL*Net to client
1146 bytes received via SQL*Net from client
68 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
2000 rows processed
n1@TEST11G> set autot off
n1@TEST11G> select 2000/30+30 from dual;
2000/30+30
----------
96.6666667
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