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a guest Mar 26th, 2019 72 Never
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  1. # 30193
  2. SELECT
  3. #   COUNT(DISTINCT pa.id)
  4.     pa.*
  5.     , ba.*
  6. FROM prices_all pa
  7. LEFT JOIN brand_groups bg
  8.               ON bg.parent_id = pa.brand_id
  9. LEFT JOIN tecdoc_2016q1_europe.ARTICLES_UPD_2017 art
  10.               ON art.ART_SUP_ID = bg.BRA_ID AND art.OWN_S_CLEAR = pa.art_num
  11. LEFT JOIN brand_all ba ON pa.brand_id = ba.id
  12. WHERE
  13.     1 = 1
  14.     AND art.ART_ID IS NOT NULL
  15.     AND pa.avalaible_gen = 1
  16.     AND art.OWN_ART_TYP = '2018'
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