How to Tune SQL with IN Subquery with Intersect for Oracle?

Here is an example SQL that retrieves data from EMPLOYEE and DEPARTMENT table with the employee’s grade code in the GRADE table.

SELECT emp_id,
       emp_name,
       dpt_name
FROM   employee,
       department
WHERE  emp_dept = dpt_id
       AND emp_grade IN (SELECT grd_id
                         FROM grade
                         WHERE grd_min_salary < 200000)
and emp_dept < ‘D’

Here the following is the query plan of this SQL, it takes 8.3 seconds to finish. The query plan shows a Hash Join with GRADE and EMPLOYEE and then hash join to DEPARTMENT. It looks like Oracle gave up any Nested Loops operations after the actual number of rows is returned from the GRADE table in this adaptive plan.

In order to ask Oracle to consider the Nested Loops operations, I added an extra Intersect operation in the subquery to rapidly narrow down the result set of grd_id returned from the GRADE table first.

SELECT emp_id,
       emp_name,
       dpt_name
FROM   employee,
       department
WHERE  emp_dept = dpt_id
       AND emp_grade IN (SELECT grd_id
                         FROM   grade
                         WHERE  grd_min_salary < 200000                          INTERSECT SELECT e1.emp_grade
                                   FROM employee e1
                                   WHERE emp_dept < ‘D’)
       AND emp_dept < ‘D’

The rewritten SQL generates a query plan that is entirely different from the original query plan, The new plan is using “Nested Loops” from DEPARTMENT to EMPLOYEE as the first steps and then Hash Join to the GRADE table. The new plan now takes 0.81 seconds only.


This kind of rewrite can be achieved by Tosska SQL Tuning Expert Pro for Oracle automatically, it shows that the rewrite is more than 10 times faster than the original SQL.

https://tosska.com/tosska-sql-tuning-expert-pro-tse-pro-for-oracle/

How to Tune SQL Statement with CASE Expression by Hints Injection for Oracle?

Here the following is a simple SQL statement with a CASE expression syntax.

SELECT *
FROM   employee
WHERE
      CASE
      WHEN emp_salary< 1000
      THEN  ‘low’
      WHEN emp_salary>100000
      THEN  ‘high’
      ELSE  ‘Normal’
      END = ‘low’

Here the following are the query plans of this SQL, it takes 4.64 seconds to finish. The query shows a Full Table Scan of the EMPLOYEE table due to the CASE expression cannot utilize the emp_salary index. It is because the CASE statement disabled the index range search of the emp_salary index.

Commonly, we will try to enable index search by forcing the SQL with an Index hint as the following:

SELECT/*+ INDEX(@SEL$1 EMPLOYEE) */ *
FROM   employee
WHERE CASE
      WHEN emp_salary < 1000
      THEN  ‘low’
      WHEN emp_salary > 100000
      THEN  ‘high’
      ELSE  ‘Normal’
     END = ‘low’

Although the CASE statement disabled the index range search of the emp_salary index, an index full scan is now enabled to help filter the result more quickly compared with the original full table scan of the EMPLOYEE table.

This hint injection takes 0.38 seconds and it is 12 times faster than the original SQL will full table scan. For this kind of SQL statement that you cannot change your source code, you can use SQL Patch with the hints and SQL text deployed to the database without the need of changing your source code.

If you can modify your source code, the best performance will be to rewrite the CASE expression into the following syntax with multiple OR conditions.

SELECT *
FROM   employee
WHERE emp_salary < 1000
     AND ‘low’ = ‘low’
     OR NOT  ( emp_salary < 1000 )
        AND  emp_salary > 100000
        AND  ‘high’ = ‘low’
     OR NOT  ( emp_salary < 1000
           OR emp_salary > 100000 )
        AND  ‘Normal’ = ‘low’

The new query plan shows an INDEX RANGE SCAN OF emp_salary index.

This kind of rewrite and hints injection can be achieved by Tosska SQL Tuning Expert Pro for Oracle automatically,

https://tosska.com/tosska-sql-tuning-expert-pro-tse-pro-for-oracle/

How to use ROWID to improve an UPDATE statement for Oracle?

Here the following is an Update SQL with a subquery that updates the EMPLOYEE table if the emp_dept satisfies the records returned from a subquery.

update  employee
   set  emp_name = ‘testing’
 where  emp_dept IN (select dpt_id
            from department
          where dpt_name like ‘A%’)
and emp_grade>2000

You can see Oracle uses a Hash join of the DEPARTMENT table and EMPLOYEE table to execute the update process. This query plan takes 1.96 seconds to complete and no index is used even though emp_dept, dpt_id, and emp_grade are indexed columns. It looks like the most expansive operation is the Table Access Full scan of the EMPLOYEE table.

Let’s rewrite the SQL into the following syntax to eliminate EMPLOYEE’s Table Access Full operation from the query plan.  The new subquery with the italic Bold text is used to force the EMPLOYEE to extract records with emp_dept in the DEPARTMENT table with the dpt_name like ‘A%’. The ROWID returned from the EMPLOYEE(subquery) is to make sure a more efficient table ROWID access to the outer EMPLOYEE table.

UPDATE  employee
SET   emp_name=‘testing’
WHERE   ROWID IN (SELECT  ROWID
          FROM   employee
          WHERE  emp_dept IN (SELECT  dpt_id
                      FROM   department
                      WHERE  dpt_name LIKE‘A%’))
     AND emp_grade > 2000

You can see the final query plan with this syntax has a better cost without full table access to the EMPLOYEE table. The new syntax takes 0.9 seconds and it is more than 2 times faster than the original syntax.

This kind of rewrite can be achieved by Tosska SQL Tuning Expert Pro for Oracle automatically, there is another SQL rewrite with similar performance, but it is not suitable to discuss in this short article, maybe I can discuss it later in my blog.

https://tosska.com/tosska-sql-tuning-expert-pro-tse-pro-for-oracle/

How to index SQL with aggregate function SQL for Oracle?

Here the following is an example SQL shows you that select the maximum emp_address which is not indexed in the EMPLOYEE table with 3 million records, the emp_grade is an indexed column.

select max(emp_address) from employee a
where emp_grade<4000

As 80% of the EMPLOYEE table’s records will be retrieved to examine the maximum emp_address string. The query plan of this SQL shows a Table Access Full on EMPLOYEE table is reasonable.

How many ways to build an index to improve this SQL?
Although it is simple SQL, there are still 3 ways to build an index to improve this SQL, the following are the possible indexes that can be built for the SQL, the first one is a single column index and the 2 and 3 are the composite index with a different order.
1. EMP_ADDRESS
2. EMP_GRADE, EMP_ADDRESS
3. EMP_ADDRESS, EMP_GRADE

Most people may use the EMP_ADDRESS as the first choice to improve this SQL, let’s see what the query plan is if we build a virtual index for the EMP_ADDRESS column in the following, you can see the estimated cost is reduced by almost half, but this query plan is finally not being used after the physical index is built for benchmarking due to actual statistics is collected.

The following query shows the EMP_ADDRESS index is not used and the query plan is the same as the original SQL without any new index built.

Let’s try the second composite index (EMP_GRADE, EMP_ADDRESS), the new query plan shows an Index Fast Full Scan of this index, it is a reasonable plan which no table’s data is needed to retrieve. So, the execution time is reduced from 16.83 seconds to 3.89 seconds.

Let’s test the last composite index (EMP_ADDRESS, EMP_GRADE) that EMP_ADDRESS is placed as the first column in the composite index, it creates a new query plan that shows an extra FIRST ROW operation for the INDEX FULL SCAN (MIN/MAX), it highly reduces the execution time from 16.83 seconds to 0.08 seconds.

So, indexing sometimes is an art that needs you to pay more attention to it, some potential solutions may perform excess your expectation.

The best index solution is now more than 200 times better than the original SQL without index, this kind of index recommendation can be achieved by Tosska SQL Tuning Expert for Oracle automatically.

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How is the order of the columns in a composite index affecting a subquery performance for Oracle?

MySQL database and sql

We know the order of the columns in a composite index will determine the usage of the index or not against a table. A query will use a composite index only if the where clause of the query has at least the leading/left-most columns of the index in it. But, it is far more complicated in correlated subquery situations. Let’s have an example SQL to elaborate the details in the following.

SELECT D.*
FROM   department D
WHERE EXISTS (SELECT    Count(*)
         FROM     employee E
         WHERE     E.emp_id < 1050000
                AND E.emp_dept = D.dpt_id
         GROUP BY  E.emp_dept
         HAVING    Count(*) > 124)

Here the following is the query plan of the SQL, it takes 10 seconds to finish. We can see that the SQL can utilize E.emp_id and E.emp_dept indexes individually.

Let’s see if a new composite index can help to improve the SQL’s performance or not, as a rule of thumb, a higher selectivity column E.emp_id will be set as the first column in a composite index (E.emp_id, E.emp_dept).

The following is the query plan of a new composite index (E.emp_id, E.emp_dept) and the result performance is not good, it takes 11.8 seconds and it is even worse than the original query plan.

If we change the order of the columns in the composite index to (E.emp_dept, E.emp_id), the following query plan is generated and the speed is improved to 0.31 seconds.

The above two query plans are similar, the only difference is the “2” operation. The first composite index with first column E.emp_id uses an INDEX RANGE SCAN of the new composite index, but the second query plan uses an INDEX SKIP SCAN for the first column of E.emp_dept composite index. You can see there is an extra filter operation for E.emp_dept in the Predicate Information of INDEX RANGE SCAN of the index (E.emp_id, E.emp_dept). But the (E.emp_dept, E.emp_id) composite index use INDEX SKIP SCAN without extra operation to filter the E.emp_dept again.

So, you have to test the order of composite index very carefully for correlated subqueries, sometimes it will give you improvements that exceed your expectation.

This kind of index recommendation can be achieved by Tosska SQL Tuning Expert for Oracle automatically.

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How to use ORDERED Hint to Tune a SQL with subquery for Oracle?

Here the following is the description of the ORDERED hint.

The ORDERED hint causes Oracle to join tables in the order in which they appear in the FROM clause.

If you omit the ORDERED hint from a SQL statement performing a join, then the optimizer chooses the order in which to join the tables. You might want to use the ORDERED hint to specify a join order if you know something about the number of rows selected from each table that the optimizer does not. Such information lets you choose an inner and outer table better than the optimizer could.

We usually use an ORDERED hint to control the john order, but how this hint causes a SQL with a subquery. Let’s use the following SQL as an example to see how ORDERED hint works for a subquery.

SELECT *
     FROM DEPARTMENT
where  dpt_id
     in (select emp_dept from employee
      where emp_id >3300000)

Here the following is the query plan of the SQL, it takes 68.84 seconds to finish. The query shows a “TABLE ACCESS FULL” of the DEPARTMENT table and “NESTED LOOPS SEMI” to an “INDEX RANGE SCAN” of EMPLOYEE.

If you think it is not an effective plan, you may want to try to reorder the join path and see if an ORDERED hint is working or not in a subquery case like this:

SELECT  /*+ ORDERED */ *
FROM  department
WHERE  dpt_id IN (SELECT  emp_dept
         FROM  employee
         WHERE  emp_id > 3300000)

Here is the query plan of the hinted SQL and the speed is 3.44 seconds which is 20 times better than the original SQL. The new query plan shows the new join order that EMPLOYEE is retrieve first and then hash join DEPARTMENT later. You can see the ORDERED hint will order the subquery’s table first. This new order clauses a new data retrieval method from the EMPLOYEE table, it makes the overall performance much better than the original query plan.

This kind of rewrite can be achieved by Tosska SQL Tuning Expert for Oracle automatically, there are other hints-injection SQL with better performance, but it is not suitable to discuss in this short article, maybe I can discuss later in my blog.

https://tosska.com/tosska-sql-tuning-expert-pro-tse-pro-for-oracle/