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Oct 24th, 2019
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  1. SELECT COUNTRY.NAME_CTR, CITY.NAME_C, ISNULL(COUNT(DISTINCT(TOURIST.ID_TRT)), 0) FROM COUNTRY INNER JOIN CITY
  2. ON COUNTRY.ID_CTR = CITY.ID_CTR LEFT JOIN HOTEL
  3. ON CITY.ID_CTY = HOTEL.ID_CTY INNER JOIN TOUR
  4. ON HOTEL.ID_H = TOUR.ID_H LEFT JOIN TOURIST
  5. ON TOUR.ID_TRT = TOURIST.ID_TRT LEFT JOIN TYPE_B
  6. ON HOTEL.ID_TB = TYPE_B.ID_TB
  7. WHERE TYPE_B.NAME_TB='BB' OR TYPE_B.NAME_TB='FB' OR TYPE_B.NAME_TB IS NULL
  8. GROUP BY COUNTRY.NAME_CTR, CITY.NAME_C
  9. HAVING COUNT(DISTINCT(TOURIST.ID_TRT)) < 2
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