比较 Oracle 中的两个连接查询
Comparing two join queries in Oracle
我有 2 个查询做同样的工作：
SELECT * FROM student_info
INNER JOIN class
ON student_info.id = class.studentId
WHERE student_info.name = 'Ken'
SELECT * FROM (SELECT * FROM student_info WHERE name = 'Ken') studInfo
INNER JOIN class
ON student_info.id = class.studentId
哪个更快？我猜是第二个但不确定，我使用的是 Oracle 11g。
更新：
我的表没有索引，我确认两个 PLAN_TABLE_OUTPUTs 几乎相同：
我肯定会倾向于第一个查询。
当select嵌套时，Oracle优化机会较少。它通常必须将内部 select 评估为临时视图，然后将外部 select 应用于该视图。这很少比 JOIN 快，其中 Oracle 将一起评估所有内容。
显示您的 EXPLAIN PLAN 也会为我们提供额外的信息。
在最新版本的 Oracle 中，优化器 足够智能，可以完成它的工作。所以没关系，您的两个查询都将在内部 优化 以有效地完成任务。优化器可能会执行 查询重写 并选择 高效执行计划 。
让我们用一个 EMP 和 DEPT 的小例子来理解这一点 table。我将在问题中使用两个与您类似的查询。
我将采用两种情况，第一种是具有非索引列的谓词，第二种是具有索引列的谓词。
情况 1 - 具有非索引列的谓词
SQL> explain plan for
2 SELECT * FROM emp e
3 INNER JOIN dept d
4 ON e.deptno = d.deptno
5 where ename = 'SCOTT';
Explained.
SQL>
SQL> SELECT * FROM TABLE(dbms_xplan.display);
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
Plan hash value: 3625962092
----------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 59 | 4 (0)| 00:00:01 |
| 1 | NESTED LOOPS | | | | | |
| 2 | NESTED LOOPS | | 1 | 59 | 4 (0)| 00:00:01 |
|* 3 | TABLE ACCESS FULL | EMP | 1 | 39 | 3 (0)| 00:00:01 |
|* 4 | INDEX UNIQUE SCAN | PK_DEPT | 1 | | 0 (0)| 00:00:01 |
| 5 | TABLE ACCESS BY INDEX ROWID| DEPT | 1 | 20 | 1 (0)| 00:00:01 |
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
3 - filter("E"."ENAME"='SCOTT')
4 - access("E"."DEPTNO"="D"."DEPTNO")
Note
-----
- this is an adaptive plan
22 rows selected.
SQL>
SQL> explain plan for
2 SELECT * FROM (SELECT * FROM emp WHERE ename = 'SCOTT') e
3 INNER JOIN dept d
4 ON e.deptno = d.deptno;
Explained.
SQL>
SQL> SELECT * FROM TABLE(dbms_xplan.display);
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
Plan hash value: 3625962092
----------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 59 | 4 (0)| 00:00:01 |
| 1 | NESTED LOOPS | | | | | |
| 2 | NESTED LOOPS | | 1 | 59 | 4 (0)| 00:00:01 |
|* 3 | TABLE ACCESS FULL | EMP | 1 | 39 | 3 (0)| 00:00:01 |
|* 4 | INDEX UNIQUE SCAN | PK_DEPT | 1 | | 0 (0)| 00:00:01 |
| 5 | TABLE ACCESS BY INDEX ROWID| DEPT | 1 | 20 | 1 (0)| 00:00:01 |
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
3 - filter("ENAME"='SCOTT')
4 - access("EMP"."DEPTNO"="D"."DEPTNO")
Note
-----
- this is an adaptive plan
22 rows selected.
SQL>
情况 2 - 具有索引列的谓词
SQL> explain plan for
2 SELECT * FROM emp e
3 INNER JOIN dept d
4 ON e.deptno = d.deptno
5 where empno = 7788;
Explained.
SQL>
SQL> SELECT * FROM TABLE(dbms_xplan.display);
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
Plan hash value: 2385808155
----------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 59 | 2 (0)| 00:00:01 |
| 1 | NESTED LOOPS | | 1 | 59 | 2 (0)| 00:00:01 |
| 2 | TABLE ACCESS BY INDEX ROWID| EMP | 1 | 39 | 1 (0)| 00:00:01 |
|* 3 | INDEX UNIQUE SCAN | PK_EMP | 1 | | 0 (0)| 00:00:01 |
| 4 | TABLE ACCESS BY INDEX ROWID| DEPT | 1 | 20 | 1 (0)| 00:00:01 |
|* 5 | INDEX UNIQUE SCAN | PK_DEPT | 1 | | 0 (0)| 00:00:01 |
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
3 - access("E"."EMPNO"=7788)
5 - access("E"."DEPTNO"="D"."DEPTNO")
18 rows selected.
SQL>
SQL> explain plan for
2 SELECT * FROM (SELECT * FROM emp where empno = 7788) e
3 INNER JOIN dept d
4 ON e.deptno = d.deptno;
Explained.
SQL>
SQL> SELECT * FROM TABLE(dbms_xplan.display);
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
Plan hash value: 2385808155
----------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 59 | 2 (0)| 00:00:01 |
| 1 | NESTED LOOPS | | 1 | 59 | 2 (0)| 00:00:01 |
| 2 | TABLE ACCESS BY INDEX ROWID| EMP | 1 | 39 | 1 (0)| 00:00:01 |
|* 3 | INDEX UNIQUE SCAN | PK_EMP | 1 | | 0 (0)| 00:00:01 |
| 4 | TABLE ACCESS BY INDEX ROWID| DEPT | 1 | 20 | 1 (0)| 00:00:01 |
|* 5 | INDEX UNIQUE SCAN | PK_DEPT | 1 | | 0 (0)| 00:00:01 |
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
3 - access("EMPNO"=7788)
5 - access("EMP"."DEPTNO"="D"."DEPTNO")
18 rows selected.
SQL>
每个案例的解释计划有什么区别吗？编号
您需要向我们展示查询计划和执行统计信息才能确定。也就是说，假设 name 被索引并且统计数据相当准确，如果这两个查询没有生成相同的计划（因此，相同的性能），我会感到震惊。对于任一查询，Oracle 都可以在评估连接之前或之后自由评估谓词，因此它不太可能在两种情况下做出不同的选择。
我有 2 个查询做同样的工作：
SELECT * FROM student_info
INNER JOIN class
ON student_info.id = class.studentId
WHERE student_info.name = 'Ken'
SELECT * FROM (SELECT * FROM student_info WHERE name = 'Ken') studInfo
INNER JOIN class
ON student_info.id = class.studentId
哪个更快？我猜是第二个但不确定，我使用的是 Oracle 11g。
更新：
我的表没有索引，我确认两个 PLAN_TABLE_OUTPUTs 几乎相同：
我肯定会倾向于第一个查询。
当select嵌套时，Oracle优化机会较少。它通常必须将内部 select 评估为临时视图，然后将外部 select 应用于该视图。这很少比 JOIN 快，其中 Oracle 将一起评估所有内容。
显示您的 EXPLAIN PLAN 也会为我们提供额外的信息。
在最新版本的 Oracle 中，优化器 足够智能，可以完成它的工作。所以没关系，您的两个查询都将在内部 优化 以有效地完成任务。优化器可能会执行 查询重写 并选择 高效执行计划 。
让我们用一个 EMP 和 DEPT 的小例子来理解这一点 table。我将在问题中使用两个与您类似的查询。
我将采用两种情况，第一种是具有非索引列的谓词，第二种是具有索引列的谓词。
情况 1 - 具有非索引列的谓词
SQL> explain plan for
2 SELECT * FROM emp e
3 INNER JOIN dept d
4 ON e.deptno = d.deptno
5 where ename = 'SCOTT';
Explained.
SQL>
SQL> SELECT * FROM TABLE(dbms_xplan.display);
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
Plan hash value: 3625962092
----------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 59 | 4 (0)| 00:00:01 |
| 1 | NESTED LOOPS | | | | | |
| 2 | NESTED LOOPS | | 1 | 59 | 4 (0)| 00:00:01 |
|* 3 | TABLE ACCESS FULL | EMP | 1 | 39 | 3 (0)| 00:00:01 |
|* 4 | INDEX UNIQUE SCAN | PK_DEPT | 1 | | 0 (0)| 00:00:01 |
| 5 | TABLE ACCESS BY INDEX ROWID| DEPT | 1 | 20 | 1 (0)| 00:00:01 |
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
3 - filter("E"."ENAME"='SCOTT')
4 - access("E"."DEPTNO"="D"."DEPTNO")
Note
-----
- this is an adaptive plan
22 rows selected.
SQL>
SQL> explain plan for
2 SELECT * FROM (SELECT * FROM emp WHERE ename = 'SCOTT') e
3 INNER JOIN dept d
4 ON e.deptno = d.deptno;
Explained.
SQL>
SQL> SELECT * FROM TABLE(dbms_xplan.display);
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
Plan hash value: 3625962092
----------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 59 | 4 (0)| 00:00:01 |
| 1 | NESTED LOOPS | | | | | |
| 2 | NESTED LOOPS | | 1 | 59 | 4 (0)| 00:00:01 |
|* 3 | TABLE ACCESS FULL | EMP | 1 | 39 | 3 (0)| 00:00:01 |
|* 4 | INDEX UNIQUE SCAN | PK_DEPT | 1 | | 0 (0)| 00:00:01 |
| 5 | TABLE ACCESS BY INDEX ROWID| DEPT | 1 | 20 | 1 (0)| 00:00:01 |
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
3 - filter("ENAME"='SCOTT')
4 - access("EMP"."DEPTNO"="D"."DEPTNO")
Note
-----
- this is an adaptive plan
22 rows selected.
SQL>
情况 2 - 具有索引列的谓词
SQL> explain plan for
2 SELECT * FROM emp e
3 INNER JOIN dept d
4 ON e.deptno = d.deptno
5 where empno = 7788;
Explained.
SQL>
SQL> SELECT * FROM TABLE(dbms_xplan.display);
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
Plan hash value: 2385808155
----------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 59 | 2 (0)| 00:00:01 |
| 1 | NESTED LOOPS | | 1 | 59 | 2 (0)| 00:00:01 |
| 2 | TABLE ACCESS BY INDEX ROWID| EMP | 1 | 39 | 1 (0)| 00:00:01 |
|* 3 | INDEX UNIQUE SCAN | PK_EMP | 1 | | 0 (0)| 00:00:01 |
| 4 | TABLE ACCESS BY INDEX ROWID| DEPT | 1 | 20 | 1 (0)| 00:00:01 |
|* 5 | INDEX UNIQUE SCAN | PK_DEPT | 1 | | 0 (0)| 00:00:01 |
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
3 - access("E"."EMPNO"=7788)
5 - access("E"."DEPTNO"="D"."DEPTNO")
18 rows selected.
SQL>
SQL> explain plan for
2 SELECT * FROM (SELECT * FROM emp where empno = 7788) e
3 INNER JOIN dept d
4 ON e.deptno = d.deptno;
Explained.
SQL>
SQL> SELECT * FROM TABLE(dbms_xplan.display);
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
Plan hash value: 2385808155
----------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 59 | 2 (0)| 00:00:01 |
| 1 | NESTED LOOPS | | 1 | 59 | 2 (0)| 00:00:01 |
| 2 | TABLE ACCESS BY INDEX ROWID| EMP | 1 | 39 | 1 (0)| 00:00:01 |
|* 3 | INDEX UNIQUE SCAN | PK_EMP | 1 | | 0 (0)| 00:00:01 |
| 4 | TABLE ACCESS BY INDEX ROWID| DEPT | 1 | 20 | 1 (0)| 00:00:01 |
|* 5 | INDEX UNIQUE SCAN | PK_DEPT | 1 | | 0 (0)| 00:00:01 |
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
3 - access("EMPNO"=7788)
5 - access("EMP"."DEPTNO"="D"."DEPTNO")
18 rows selected.
SQL>
每个案例的解释计划有什么区别吗？编号
您需要向我们展示查询计划和执行统计信息才能确定。也就是说，假设 name 被索引并且统计数据相当准确，如果这两个查询没有生成相同的计划（因此，相同的性能），我会感到震惊。对于任一查询，Oracle 都可以在评估连接之前或之后自由评估谓词，因此它不太可能在两种情况下做出不同的选择。