[Solved] sql subqueries explained

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sql subqueries explained

SELECT sub.*
FROM (
     SELECT *
     FROM tutorial.sf_crime_incidents_2014_01
     WHERE day_of_week = 'Friday'
     ) sub
WHERE sub.resolution = 'NONE'

This subquery is not mine, it is from a tutorial website. I just want to clear the idea of the subquery, so first, the inner query is done and then the answer becomes in the alias name <code>sub</code>. Then the outer query takes the answer and selects all columns matching this answer with a where condition.Is this how it works?

Thank you for the help.

How to optimize this SQL query?

The following recommendations will help you in your SQL tuning process.
You'll find 3 sections below:

  1. Description of the steps you can take to speed up the query.
  2. The optimal indexes for this query, which you can copy and create in your database.
  3. An automatically re-written query you can copy and execute in your database.
The optimization process and recommendations:
  1. Avoid Selecting Unnecessary Columns (query line: 2): Avoid selecting all columns with the '*' wildcard, unless you intend to use them all. Selecting redundant columns may result in unnecessary performance degradation.
  2. Create Optimal Indexes (modified query below): The recommended indexes are an integral part of this optimization effort and should be created before testing the execution duration of the optimized query.
  3. Prefer Direct Join Over Joined Subquery (query line: 4): We advise against using subqueries as they are not optimized well by the optimizer. Therefore, we recommend to replace subqueries with JOIN clauses.
Optimal indexes for this query:
ALTER TABLE `sf_crime_incidents_2014_01` ADD INDEX `sf_incidents_idx_resolution_day_week` (`resolution`,`day_of_week`);
The optimized query:
SELECT
        sub.* 
    FROM
        tutorial.sf_crime_incidents_2014_01 sub 
    WHERE
        (
            sub.resolution = 'NONE'
        ) 
        AND (
            sub.day_of_week = 'Friday'
        )

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* original question posted on StackOverflow here.