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Aug 23rd, 2019
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  1. select
  2. userid,
  3. avg(duration) avg_duration
  4. from
  5. sessions x
  6. where
  7. exists (
  8. select 1
  9. from
  10. sessions y
  11. where
  12. y.userid = x.userid
  13. and y.id <> x.id
  14. )
  15. group by
  16. userid;
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