[Solved] SQL - Multiple many-to-many relations filtering SELECT

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 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. 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 `Cadastros` ADD INDEX `cadastros_idx_id` (`id`);
ALTER TABLE `Convenios` ADD INDEX `convenios_idx_id` (`id`);
ALTER TABLE `Especialidades` ADD INDEX `especialidades_idx_id` (`id`);
ALTER TABLE `Facilidades` ADD INDEX `facilidades_idx_id` (`id`);
ALTER TABLE `cadastros_convenios` ADD INDEX `cadastros_convenio_idx_cadastro_id` (`cadastro_id`);
ALTER TABLE `cadastros_especialidades` ADD INDEX `cadastros_especial_idx_cadastro_id` (`cadastro_id`);
ALTER TABLE `cadastros_facilidades` ADD INDEX `cadastros_facilida_idx_cadastro_id` (`cadastro_id`);
The optimized query:
SELECT
        Cadastro.*,
        Convenio.* 
    FROM
        Cadastros AS Cadastro 
    INNER JOIN
        cadastros_convenios AS CadastrosConvenio 
            ON (
                Cadastro.id = CadastrosConvenio.cadastro_id
            ) 
    INNER JOIN
        Convenios AS Convenio 
            ON (
                CadastrosConvenio.convenio_id = Convenio.id 
                AND Convenio.id IN (
                    2,
                3)) 
            INNER JOIN
                cadastros_especialidades AS CadastrosEspecialidade 
                    ON (
                        Cadastro.id = CadastrosEspecialidade.cadastro_id
                    ) 
            INNER JOIN
                Especialidades AS Especialidade 
                    ON (
                        CadastrosEspecialidade.especialidade_id = Especialidade.id 
                        AND Especialidade.id IN (
                            1
                        )
                    ) 
            INNER JOIN
                cadastros_facilidades AS CadastrosFacilidade 
                    ON (
                        Cadastro.id = CadastrosFacilidade.cadastro_id
                    ) 
            INNER JOIN
                Facilidades AS Facilidade 
                    ON (
                        CadastrosFacilidade.facilidade_id = Facilidade.id 
                        AND Facilidade.id IN (
                            1,
                        2)) 
                    GROUP BY
                        Cadastro.id 
                    HAVING
                        COUNT(*) = 5 
                    ORDER BY
                        NULL

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* original question posted on StackOverflow here.