Question

我在使用不同的正确查询更改PL / SQL oracle中的查询时遇到问题。当前查询：

SELECT MAX (workzone)
           FROM sccd_device_uim_tab
          WHERE NAME LIKE 18075009 || '%';

我的客户需要修改查询，因为：

Full table with a % should definitely be avoided.

这个问题的解决方案是什么？

提前致谢

从包中添加查询

PROCEDURE sccd_get_impactservice_manual (
      in_faultid          IN       VARCHAR2,
      in_deviceid         IN       VARCHAR2,
      in_status           IN       VARCHAR2,
      in_opendate         IN       DATE,
      in_closedate        IN       DATE,
      out_impact_result   OUT      tcur,
      out_count_service   OUT      NUMBER,
      out_workzone        OUT      VARCHAR2,
      p_ret_char          OUT      VARCHAR2,
      p_ret_number        OUT      NUMBER,
      p_ret_msg           OUT      VARCHAR2
   )
   IS
   BEGIN
      SELECT orauser.ossa_get_sto_from_device (in_deviceid)
        INTO out_workzone
        FROM DUAL;

      IF out_workzone IS NULL
      THEN
         SELECT MAX (workzone)
           INTO out_workzone
           FROM sccd_device_uim_tab
          WHERE NAME = in_deviceid; --//this is I try to solve
          --WHERE NAME LIKE in_deviceid || '%'; //this is the current issue
      END IF;

Answer 1

在哪里名称像in_deviceid || '％';

绝对应该避免使用带有％的全表。

你错了。当您使用 LIKE 运算符作为'STRING-%'时，如果列上有任何索引，Oracle将执行 INDEX SCAN 。使用统计信息更新，您不应该看到全表扫描。

让我们看一个测试用例：

SQL> CREATE TABLE emp_new AS SELECT ename FROM emp;

Table created.

SQL>
SQL> EXEC DBMS_STATS.gather_table_stats('LALIT', 'EMP_NEW');

PL/SQL procedure successfully completed.

SQL>

LIKE ename ||'％'：没有索引：

SQL> EXPLAIN PLAN FOR
  2  SELECT * FROM emp_new WHERE ename LIKE ename||'%';

Explained.

SQL>
SQL> SELECT * FROM TABLE(dbms_xplan.display);

PLAN_TABLE_OUTPUT
-----------------------------------------------------------------------------
Plan hash value: 565523140

-----------------------------------------------------------------------------
| Id  | Operation         | Name    | Rows  | Bytes | Cost (%CPU)| Time     |
-----------------------------------------------------------------------------
|   0 | SELECT STATEMENT  |         |     2 |    12 |     3   (0)| 00:00:01 |
|*  1 |  TABLE ACCESS FULL| EMP_NEW |     2 |    12 |     3   (0)| 00:00:01 |
-----------------------------------------------------------------------------

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

PLAN_TABLE_OUTPUT
-----------------------------------------------------------------------------

   1 - filter("ENAME" LIKE "ENAME"||'%')

13 rows selected.

SQL>

因此，正如预期的那样，Oracle会进行全表扫描。

LIKE ename ||'％'：WITH Index：

SQL> CREATE INDEX idx_nm ON emp_new(ename);

Index created.

SQL>
SQL> EXEC DBMS_STATS.gather_table_stats('LALIT', 'EMP_NEW');

PL/SQL procedure successfully completed.

SQL>
SQL> EXPLAIN PLAN FOR
  2  SELECT * FROM emp_new WHERE ename LIKE ename||'%';

Explained.

SQL>
SQL> SELECT * FROM TABLE(dbms_xplan.display);

PLAN_TABLE_OUTPUT
---------------------------------------------------------------------------
Plan hash value: 848277793

---------------------------------------------------------------------------
| Id  | Operation        | Name   | Rows  | Bytes | Cost (%CPU)| Time     |
---------------------------------------------------------------------------
|   0 | SELECT STATEMENT |        |     2 |    12 |     1   (0)| 00:00:01 |
|*  1 |  INDEX FULL SCAN | IDX_NM |     2 |    12 |     1   (0)| 00:00:01 |
---------------------------------------------------------------------------

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

PLAN_TABLE_OUTPUT
---------------------------------------------------------------------------

   1 - filter("ENAME" LIKE "ENAME"||'%')

13 rows selected.

SQL>

因此，您可以清楚地看到正在使用索引。

如何使用LIKE运算符避免全表扫描

1 个答案: