[Solved] Select the first image from multiple images using JOIN mysql

How to optimize this SQL query?

In case you have your own slow SQL query, you can optimize it automatically here.

For the query above, the following recommendations will be helpful as part of the 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 Subqueries (query line: 13): We advise against using subqueries as they are not optimized well by the optimizer. Therefore, it's recommended to join a newly created temporary table that holds the data, which also includes the relevant search index.
  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. Explicitly ORDER BY After GROUP BY (modified query below): By default, the database sorts all 'GROUP BY col1, col2, ...' queries as if you specified 'ORDER BY col1, col2, ...' in the query as well. If a query includes a GROUP BY clause but you want to avoid the overhead of sorting the result, you can suppress sorting by specifying 'ORDER BY NULL'.
Optimal indexes for this query:
ALTER TABLE `fav_food` ADD INDEX `fav_food_idx_user_id` (`user_id`);
ALTER TABLE `food` ADD INDEX `food_idx_f_id` (`f_id`);
ALTER TABLE `food_image` ADD INDEX `food_image_idx_rank` (`rank`);
ALTER TABLE `food_image` ADD INDEX `food_image_idx_f_id_rank` (`f_id`,`rank`);
The optimized query:
SELECT
        food.f_id,
        food.description,
        img.minimgrank,
        i.img_url AS profile_photo 
    FROM
        fav_food 
    INNER JOIN
        food 
            ON fav_food.f_id = food.f_id 
    LEFT OUTER JOIN
        (
            SELECT
                food_image.f_id,
                Min(food_image.rank) AS minImgRank 
            FROM
                food_image 
            GROUP BY
                food_image.f_id 
            ORDER BY
                NULL
        ) img 
            ON img.f_id = food.f_id 
    JOIN
        food_image i 
            ON i.rank = img.minimgrank 
    WHERE
        fav_food.user_id = ?

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* original question posted on StackOverflow here.