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Home/ Questions/Q 7192717
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T20:00:00+00:00 2026-05-28T20:00:00+00:00

I’m wondering if there’s any way to optimize the following SELECT query. (Note: I

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I’m wondering if there’s any way to optimize the following SELECT query. (Note: I typed this when writing my question for nonexistent tables and I might not have the correct syntax.)

The goal is, if Table2 contains any related rows I want to set the value of the third column to the number of related rows in Table2. Otherwise, if Table3 contains any related rows I want to set the column to the number of related rows in Table3. Otherwise, I want to set the column value to 0.

SELECT Id, Title,
    CASE
        WHEN EXISTS (SELECT * FROM Table2 t2 WHERE t2.RelatedId = Table1.Id) THEN
            (SELECT COUNT(1) FROM Table2 t2 WHERE t2.RelatedId = Table1.Id)
        WHEN EXISTS (SELECT * FROM Table3 t3 WHERE t3.RelatedId = Table1.Id) THEN
            (SELECT COUNT(1) FROM Table3 t3 WHERE t3.RelatedId = Table1.Id)
        ELSE 0
    END AS RelatedCount
    FROM Table1

I don’t like the fact that I’m basically performing the same query twice (in two cases). Is there any way to do what I want while only performing the query once?

Note that this is part of a much larger query with multiple JOINs and UNIONs so it’s not easy to take a completely different approach.

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  1. Editorial Team
    Editorial Team
    2026-05-28T20:00:01+00:00Added an answer on May 28, 2026 at 8:00 pm

    This query should perform much better. You are not just performing the same query twice; since they are correlated subqueries, they will run once per row.

    SELECT Id, Title,
        coalesce(t2.Count, t3.Count, 0) AS RelatedCount
        FROM Table1 t
    left outer join (
        SELECT RelatedId, count(*) as Count
        FROM Table2
        group by RelatedId
    ) t2 on t1.Id = t2.RelatedId
    left outer join (
        SELECT RelatedId, count(*) as Count
        FROM Table3
        group by RelatedId
    ) t3 on t1.Id = t3.RelatedId
    
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