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Oct 5th, 2015
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  1.  
  2.  
  3. select b.id as genre_id, b.name as genre_name from
  4. (select distinct category_id from subcategory_relationships where parent_category_id = '5601d06e95023a0a2286ae9c') a
  5. LEFT OUTER JOIN
  6. (select id, name from categories) b
  7. ON (a.category_id = b.id)
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