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Old 12-17-2012, 01:05 AM   PM User | #2
Old Pedant
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Code:
SELECT P1.some, P1.fields
FROM patterns AS P1, 
     ( SELECT lnktxt, category, COUNT(*) AS howmany
      FROM patterns 
      GROUP BY linktxt, category
      HAVING howmany > 1 ) AS P2
WHERE P1.lnktxt = P2.lnktxt AND P1.category = P2.category
???

If I understood what you were saying.
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