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    <p>Unfortunately, MySQL is not very good at optimizing subqueries with IN. This is from <a href="http://dev.mysql.com/doc/refman/5.5/en/subquery-restrictions.html" rel="nofollow">MySQL documentation</a>:</p> <blockquote> <p>Subquery optimization for IN is not as effective as for the = operator or for the IN(value_list) operator.</p> <p>A typical case for poor IN subquery performance is when the subquery returns a small number of rows but the outer query returns a large number of rows to be compared to the subquery result.</p> <p>The problem is that, for a statement that uses an IN subquery, the optimizer rewrites it as a correlated subquery. Consider the following statement that uses an uncorrelated subquery:</p> <p>SELECT ... FROM t1 WHERE t1.a IN (SELECT b FROM t2);</p> <p>The optimizer rewrites the statement to a correlated subquery:</p> <p>SELECT ... FROM t1 WHERE EXISTS (SELECT 1 FROM t2 WHERE t2.b = t1.a);</p> <p>If the inner and outer queries return M and N rows, respectively, the execution time becomes on the order of O(M×N), rather than O(M+N) as it would be for an uncorrelated subquery.</p> <p>An implication is that an IN subquery can be much slower than a query written using an IN(value_list) operator that lists the same values that the subquery would return.</p> </blockquote> <p>Try using a JOIN instead.</p> <p>Because MySQL works from the inside out, sometimes you can trick MySQL by wrapping the subquery inside yet another subquery like so:</p> <pre><code>SELECT COUNT(*) FROM table_name WHERE device_id IN (SELECT * FROM (SELECT DISTINCT device_id FROM table_name WHERE NAME = 'SOME_PARA') tmp) </code></pre> <p>Here's the JOIN solution:</p> <pre><code>SELECT COUNT(DISTINCT t2.id) FROM table_name t1 JOIN table_name t2 ON t2.device_id = t1.device_id WHERE t1.NAME = 'SOME_PARA' </code></pre> <p>Notice that I start from the inside and go out also.</p>
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    1. PODoes MySQL optimize subqueries automatically?
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    1. USMarcus Adams
    1. USMarcus Adams
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    1. COThe answer to your final question is "no". The first is counting the number of rows that have the name "SOME_PARA". The second is counting the number of rows that share a device with something called "SOME_PARA".
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      1. USGordon Linoff
    2. CO@Gordon, thanks for that. It must be my weakness. I think I have the proper JOIN solution. Feel free to edit if I got it wrong.
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      1. USMarcus Adams
    3. COStill doesn't work. You are going to multiply the number of rows. As a hint, to get the equivalent of an "IN" you need to have a "DISTINCT" for the second table.
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      1. USGordon Linoff
 

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