alter index coalesce和alter index rebuild的区别

 

alter index coalesce和alter index rebuild的区别：
alter index coalesce是合并同一branch的leaf block。
而alter index rebuild是重新建立一个indexes，一般需要2倍的index大小的空间。而且需要排序。

 

补充：

它讲到”coalesce index”比”rebuild index”消耗更少的资源。在我的测试中，coalesce index比rebuild index产生更多的redo size；如果对一些GB级别的索引进行coalesce，过多的redo会严重影响系统性能，且会很快把归档目录填满。还是在产品库上少用的为好。

以下是证明alter index coalesce的过程：
×××××××× ×××××××××××××××××alter index coalesce××××××××××××××××××××××
SQL>
SQL> select * from v$version;

BANNER
--------------------------------------------------------------------------------

Oracle9i Enterprise Edition Release 9.2.0.6.0 - 64bit Production
PL/SQL Release 9.2.0.6.0 - Production
CORE    9.2.0.6.0       Production
TNS for Solaris: Version 9.2.0.6.0 - Production
NLSRTL Version 9.2.0.6.0 - Production

SQL> drop table t2;

Table dropped.

SQL> create table t2(col1 number);

Table created.

SQL> create unique index t2_inx on t2(col1);

Index created.

SQL> alter table t2 add constraint t2_pk primary key(col1);

Table altered.

SQL> begin
  2  for i in  500001..601000 loop
  3      insert into t2 values(i);
  4  end loop;
  5  end;
  6  /

PL/SQL procedure successfully completed.

SQL> commit;

Commit complete.

SQL> begin
  2  sys.dbms_stats.gather_table_stats('SYSTEM', 'T2', ESTIMATE_PERCENT => 20, c
ascade=> true);
  3  end;
  4  /

PL/SQL procedure successfully completed.

SQL> col index_name for a10;
SQL> select index_name, blevel, leaf_blockS, distinct_keys
  2       clustering_factor, num_rows
  3       from user_indexes
  4       where table_name LIKE 'T2%';

INDEX_NAME     BLEVEL LEAF_BLOCKS CLUSTERING_FACTOR   NUM_ROWS
---------- ---------- ----------- ----------------- ----------
T2_INX              1         190            101000     101000

SQL> declare
  2  cursor mm is select col1 from t2;
  3  begin
  4  for cur_1 in mm loop
  5     if mod(cur_1.col1, 2) = 0 then
  6       delete from t2 where col1=cur_1.col1;
  7     end if;
  8  end loop;
  9  commit;
 10  end;
 11  /

PL/SQL procedure successfully completed.

SQL> begin
  2  sys.dbms_stats.gather_table_stats('SYSTEM', 'T2', ESTIMATE_PERCENT => 20, c
ascade=> true);
  3  end;
  4  /

PL/SQL procedure successfully completed.

SQL> select index_name, blevel, leaf_blockS, distinct_keys
  2       clustering_factor, num_rows
  3       from user_indexes
  4       where table_name LIKE 'T2%';

INDEX_NAME     BLEVEL LEAF_BLOCKS CLUSTERING_FACTOR   NUM_ROWS
---------- ---------- ----------- ----------------- ----------
T2_INX              1         190             50500      50500

SQL> alter index t2_inx coalesce;

Index altered.

SQL> begin
  2  sys.dbms_stats.gather_table_stats('SYSTEM', 'T2', ESTIMATE_PERCENT => 20, c
ascade=> true);
  3  end;
  4  /

PL/SQL procedure successfully completed.

SQL> select index_name, blevel, leaf_blockS, distinct_keys
  2       clustering_factor, num_rows
  3       from user_indexes
  4       where table_name LIKE 'T2%';

INDEX_NAME     BLEVEL LEAF_BLOCKS CLUSTERING_FACTOR   NUM_ROWS
---------- ---------- ----------- ----------------- ----------
T2_INX              1         189             50500      50500

SQL> declare
  2  cursor mm is select col1 from t2;
  3  begin
  4  for cur_1 in mm loop
  5     if mod(cur_1.col1, 3) = 0 then
  6       delete from t2 where col1=cur_1.col1;
  7     end if;
  8  end loop;
  9  commit;
 10  end;
 11  /

PL/SQL procedure successfully completed.

SQL> alter index t2_inx coalesce;

Index altered.

SQL> begin
  2  sys.dbms_stats.gather_table_stats('SYSTEM', 'T2', ESTIMATE_PERCENT => 20, c
ascade=> true);
  3  end;
  4  /

PL/SQL procedure successfully completed.

SQL> select index_name, blevel, leaf_blockS, distinct_keys
  2       clustering_factor, num_rows
  3       from user_indexes
  4       where table_name LIKE 'T2%';

INDEX_NAME     BLEVEL LEAF_BLOCKS CLUSTERING_FACTOR   NUM_ROWS
---------- ---------- ----------- ----------------- ----------
T2_INX              1          95             33666      33666

SQL>

×××××××× ×××××××××××××××××alter index coalesce××××××××××××××××××××××

 

 

 redo size 检验

 

它讲到”coalesce index”比”rebuild index”消耗更少的资源。在我的测试中，coalesce index比rebuild index产生更多的redo size；如果对一些GB级别的索引进行coalesce，过多的redo会严重影响系统性能，且会很快把归档目录填满。还是在产品库上少用的为好。



先在一个7M的表上进行测试

PHP code:

create table binzhang(id number not null,creation date not null,last_modified date not null);

insert into binzhang select binzhang_seq.nextval id,created,created from dba_objects

insert into binzhang select binzhang_seq.nextval id,created,created from dba_objects;

Commit

SQL> select bytes from user_segments where segment_name=’BINZHANG’;

BYTES

———-

7340032

create index binzhangidx1 on binzhang(creation) ;

create index binzhangidx2 on binzhang(last_modified) ;

update binzhang set creation=creation+124,last_modified=last_modified+124 where mod(id,12)in(1,3,5,7,9,11);

commit;

analyze index binzhangidx1 validate structure;

SQL> select name,value from v$statname s,v$mystat v where s.STATISTIC#=v.STATISTIC# and name=’redo size’;

NAME VALUE

—————————–

redo size 1288055828



SQL> alter index binzhangidx1 coalesce;

Index altered.



SQL> select name,value from v$statname s,v$mystat v where s.STATISTIC#=v.STATISTIC# and name=’redo size’;

NAME VALUE

———————— ———-

redo size 1294815880



SQL> alter index binzhangidx2 rebuild tablespace cr_data;

Index altered.



SQL> select name,value from v$statname s,v$mystat v where s.STATISTIC#=v.STATISTIC# and name=’redo size’;

NAME VALUE

———————— ———-

redo size 1299289400



Redo size used by coalesce 1294815880-1288055828=6760052

Redo size used by rebuild 1299289400-1294815880=4473520



OK. We can see that coalesce index generate more redo than rebuildfora 7M table.

...............................

再在一个200M的表上进行测试。

PHP code:

SQL> select bytes from user_segments where segment_name=’BINZHANG’;

BYTES

———-

201326592

SQL> update binzhang set creation=creation+124,last_modified=last_modified+124 where mod(id,12)in(1,3,5,7,9,11);

3083428 rows updated.

SQL> commit;

SQL> select name,value from v$statname s,v$mystat v where s.STATISTIC#=v.STATISTIC# and name=’redo size’;

NAME VALUE

——————————- ———-

redo size 568



SQL> alter index binzhangidx2 rebuild;

Index altered.



SQL> select name,value from v$statname s,v$mystat v where s.STATISTIC#=v.STATISTIC# and name=’redo size’;

NAME VALUE

————————— ———-

redo size 134919152



SQL> alter index binzhangidx1 coalesce;

Index altered.



SQL> select name,value from v$statname s,v$mystat v where s.STATISTIC#=v.STATISTIC# and name=’redo size’;



NAME VALUE

———————— ———-

redo size 496401172



............................................

非常明显，coalesce产生过多的redo size.

结论：coalesce index is more resource intensive than rebuild index.