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Apr 1st, 2020
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MySQL 0.22 KB | None | 0 0
  1. select D2.Nome, D2.Cognome, D2.Dipartimento
  2. from
  3.         )
  4.         select D.Cognome
  5.         from docenti D
  6.         group by D.Cognome
  7.         having count(*) = 2
  8.         ) AS D
  9.         NATURAL JOIN
  10.         docenti D2
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