The jOOQ User Manual : SQL building : QueryParts : SQL transformation : Pattern based transformation : COUNT(*) scalar subquery comparison

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COUNT(*) scalar subquery comparison

Applies to ✅ Open Source Edition   ✅ Express Edition   ✅ Professional Edition   ✅ Enterprise Edition

When comparing a scalar subquery that calculates COUNT(*) with a single value, then chances are that weaker optimisers might be better off with an equivalent EXISTS predicate as can be seen in this blog post about COUNT(*) vs EXISTS.

This transformation is only applied under certain circumstances, including:

• In the absence of UNION and other set operations
• In the absence of GROUP BY and HAVING
• Only with COUNT(*), not with COUNT(expr)

Using Settings.transformPatternsScalarSubqueryCountAsteriskGtZero, the following transformations can be achieved:

-- With Settings.transformPatternsScalarSubqueryCountAsteriskGtZero active, this:
SELECT
  (SELECT COUNT(*) FROM tab) > 0
;

-- ... is transformed into the equivalent expression:
SELECT
  EXISTS (SELECT 1 FROM tab) -- (SELECT COUNT(*) FROM tab) > 0
;

The jOOQ User Manual : SQL building : QueryParts : SQL transformation : Pattern based transformation : COUNT(*) scalar subquery comparison

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