Question

创建一个测试表：

create table customer (first_name varchar2(20), last_name varchar2(20) not null, address varchar(20));

insert into customer select dbms_random.string('U', 20), dbms_random.string('U', 20), dbms_random.string('U', 20) from dual connect by level <= 100000;
commit;

create index i_ln_fn_0 on customer(last_name, first_name,0); — just to be sure that all rows are indexed

现在解释计划：

explain plan for
select /*+ FIRST_ROWS(20) */ *
  from CUSTOMER
  where first_name like 'AB%'
    and first_name is not null
  order by last_name;

select * from table(dbms_xplan.display);

-------------------------------------------------------------------------------
| Id  | Operation          | Name     | Rows  | Bytes | Cost (%CPU)| Time     |
-------------------------------------------------------------------------------
|   0 | SELECT STATEMENT   |          |   197 | 12411 |   275   (2)| 00:00:04 |
|   1 |  SORT ORDER BY     |          |   197 | 12411 |   275   (2)| 00:00:04 |
|*  2 |   TABLE ACCESS FULL| CUSTOMER |   197 | 12411 |   274   (1)| 00:00:04 |
-------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   2 - filter("FIRST_NAME" LIKE 'AB%' AND "FIRST_NAME" IS NOT NULL)

但是因为我只想要第一行，所以我想避免整个表格。我想有这样的计划：

SELECT STATEMENT
    TABLE ACCESS BY ROWID (customer)
        INDEX FULL SCAN (i_ln_fn_0)

如何说服数据库以避免排序？

问题更严重。即使我只使用last_name，无处不在：

explain plan for
select /*+ FIRST_ROWS(20) */ last_name
  from CUSTOMER
  where last_name like 'AB%'
    and last_name is not null
  order by last_name;

select * from table(dbms_xplan.display);

-------------------------------------------------------------------------------
| Id  | Operation         | Name      | Rows  | Bytes | Cost (%CPU)| Time     |
-------------------------------------------------------------------------------
|   0 | SELECT STATEMENT  |           |    17 |   357 |     4  (25)| 00:00:01 |
|   1 |  SORT ORDER BY    |           |    17 |   357 |     4  (25)| 00:00:01 |
|*  2 |   INDEX RANGE SCAN| I_LN_FN_0 |    17 |   357 |     3   (0)| 00:00:01 |
-------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   2 - access("LAST_NAME" LIKE 'AB%')
       filter("LAST_NAME" LIKE 'AB%')

这里排序实际上没有必要，但db仍然使用它。为什么呢？

编辑：在

上进行测试

Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - 64bit Production
PL/SQL Release 11.2.0.1.0 - Production
"CORE 11.2.0.1.0 Production"
TNS for 64-bit Windows: Version 11.2.0.1.0 - Production
NLSRTL Version 11.2.0.1.0 - Production

和

Oracle Database 11g Enterprise Edition Release 11.2.0.2.0 - 64bit Production+
PL/SQL Release 11.2.0.2.0 - Production+
"CORE 11.2.0.2.0Production"+
TNS for IBM/AIX RISC System/6000: Version 11.2.0.2.0 - Production+
NLSRTL Version 11.2.0.2.0 - Production+

采用相同的计划。

Answer 1

原因是NLS_SORT参数值错误。将其改为BINARY后，计划开始看起来像我想要的那样。

来自http://docs.oracle.com/cd/E24693_01/server.11203/e24448/initparams152.htm：

NLS_SORT的值会影响查询的执行计划。因为一个标准索引不能用作在a中排序的值的来源语言顺序，通常必须执行显式排序操作而不是索引范围扫描。 NLSSORT的功能索引可以定义函数以提供以语言顺序排序的值并将索引范围扫描重新引入执行计划。

（我在forums.oracle com中得到了Paul Horth的回答。）

使用复合索引时如何避免排序？

1 个答案: