1.物理讀(physical read)

當數據塊第一次讀取到,就會緩存到buffer cache 中,而第二次讀取和修改該數據塊時就在內存buffer cache 了 以下是例子:

1.1? 第一次讀取:

C:\Documents and Settings\Paul Yi>sqlplus "/as sysdba"

SQL*Plus: Release 9.2.0.4.0 - Production on Thu Feb 28 09:32:04?2008

Copyright (c) 1982, 2002, Oracle Corporation.? All rights reserved.

Connected to:
Oracle9i Enterprise Edition Release 9.2.0.4.0 - Production
With the Partitioning, OLAP and Oracle Data Mining options
JServer Release 9.2.0.4.0 - Production

SQL> set autotrace traceonly
SQL> select * from test;

Execution Plan
----------------------------------------------------------
?? 0????? SELECT STATEMENT ptimizer=CHOOSE (Cost=2 Card=4 Bytes=8)
?? 1??? 0?? TABLE ACCESS (FULL) OF 'TEST' (Cost=2 Card=4 Bytes=8)

?

Statistics
----------------------------------------------------------
??????? 175? recursive calls
????????? 0? db block gets
???????? 24? consistent gets
?????????9? physical reads???????????? --9個物理讀
????????? 0? redo size
??????? 373? bytes sent via SQL*Net to?client
??????? 503? bytes received via SQL*Net from client
????????? 2? SQL*Net roundtrips to/from client
????????? 2? sorts (memory)
????????? 0? sorts (disk)
????????? 1? rows processed

1.2? 第二次讀取

SQL> select * from test;

Execution Plan
----------------------------------------------------------
?? 0????? SELECT STATEMENT ptimizer=CHOOSE (Cost=2 Card=4 Bytes=8)
?? 1??? 0?? TABLE ACCESS (FULL) OF 'TEST' (Cost=2 Card=4 Bytes=8)

?

Statistics
----------------------------------------------------------
????????? 0? recursive calls
????????? 0? db block gets
????????? 7? consistent gets
????????? 0? physical reads???? --沒有發(fā)生物理讀了,直接從buffer cache 中讀取了
????????? 0? redo size
??????? 373? bytes sent via SQL*Net to client
??????? 503? bytes received via SQL*Net from client
????????? 2? SQL*Net roundtrips to/from client
????????? 0? sorts (memory)
????????? 0? sorts (disk)
????????? 1? rows processed

1.3? 數據塊被重新讀入buffer cache ,這種發(fā)生在

如果有新的數據需要被讀入Buffer Cache中，而Buffer Cache又沒有足夠的空閑空間，Oracle就根據LRU算法將LRU鏈表中LRU端的數據置換出去。當這些數據被再次訪問到時，需要重新從磁盤讀入。

SQL> alter session set events 'immediate trace name flush_cache';--清空數據緩沖區(qū)

Session altered.

SQL> select * from test;

Execution Plan
----------------------------------------------------------
?? 0????? SELECT STATEMENT ptimizer=CHOOSE (Cost=2 Card=4 Bytes=8)
?? 1??? 0?? TABLE ACCESS (FULL) OF 'TEST' (Cost=2 Card=4 Bytes=8)

?

Statistics
----------------------------------------------------------
????????? 0? recursive calls
????????? 0? db block gets
????????? 7? consistent gets
????????? 6? physical reads?? --又重新發(fā)生了物理讀
????????? 0? redo size
??????? 373? bytes sent via SQL*Net to client
??????? 503? bytes received via SQL*Net from client
????????? 2? SQL*Net roundtrips to/from client
????????? 0? sorts (memory)
????????? 0? sorts (disk)
????????? 1? rows processed

2.邏輯讀(buffer read)

邏輯讀指的就是從（或者視圖從）Buffer Cache中讀取數據塊。按照訪問數據塊的模式不同，可以分為即時讀（Current Read）和一致性讀（Consistent Read）。注意：邏輯IO只有邏輯讀，沒有邏輯寫。

• 即時讀

即時讀即讀取數據塊當前的最新數據。任何時候在Buffer Cache中都只有一份當前數據塊。即時讀通常發(fā)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。

SQL> select * from test for update;

Execution Plan
----------------------------------------------------------
?? 0????? SELECT STATEMENT ptimizer=CHOOSE (Cost=2 Card=4 Bytes=8)
?? 1??? 0?? FOR UPDATE
?? 2??? 1???? TABLE ACCESS (FULL) OF 'TEST' (Cost=2 Card=4 Bytes=8)

?

Statistics
----------------------------------------------------------
????????? 0? recursive calls
?????????1? db block gets
???????? 14? consistent gets
????????? 0? physical reads
??????? 252? redo size
??????? 386? bytes sent via SQL*Net to client
??????? 503? bytes received via SQL*Net from client
????????? 2? SQL*Net roundtrips to/from client
????????? 0? sorts (memory)
????????? 0? sorts (disk)
????????? 1? rows processed

SQL>

• 一致性讀

Oracle是一個多用戶系統(tǒng)。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。

下面這個例子幫助大家理解一下一致性讀：

會話1中：

SQL> select * from test;

??????? ID
??????? ----------
????????? 1000

SQL> update test set id=2000;

1 row updated.

會話2中:

SQL> set autotrace on
?? SQL> select * from test;

??????? ID
?????? ----------
????????? 1000

Execution Plan
----------------------------------------------------------
?? 0????? SELECT STATEMENT ptimizer=CHOOSE (Cost=2 Card=4 Bytes=8)
?? 1??? 0?? TABLE ACCESS (FULL) OF 'TEST' (Cost=2 Card=4 Bytes=8)

?

Statistics
----------------------------------------------------------
????????? 0? recursive calls
????????? 0? db block gets
?????????9? consistent gets??? 沒有事物做update時 是 7 consistent gets?說明多了2個? consistent gets????這2個是要從回滾段中獲取的?????
????????? 0? physical reads
???????? 52? redo size
??????? 373? bytes sent via SQL*Net to client
??????? 503? bytes received via SQL*Net from client
????????? 2? SQL*Net roundtrips to/from client
????????? 0? sorts (memory)
????????? 0? sorts (disk)
????????? 1? rows processed

SQL>

?

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