Latest X for each Y in a many-to-many relationship

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It can be done in one. This is an example of the "Latest X for each Y" common SQL question. See SQL FAQ: Common SQL Questions, part 3 (5 July, 2007). However, this one's a bit trickier since it involves a many-to-many join.

select I.industry
     , A.title
     , A.pubdate 
  from industries as I
inner
  join industryarticles as IA
    on IA.industry_ID = I.ID
inner
  join articles as A
    on A.ID = IA.article_ID
   and 2 >
       ( select count(*)
           from industryarticles as IA2
         inner
           join articles as A2
             on A2.ID = IA2.article_ID
          where IA2.industry_ID = I.ID
            and A2.pubdate > A.pubdate ) 
order
    by I.industry
     , A.pubdate desc

The subquery—which is correlated to the outer query using the I correlation variable to link industries and the A correlation variable to link articles—counts the number of articles in the same industry which have a later date than the article in the outer query. This count must be less than the number 2. In other words, it must be 1 or 0. If the article being considered in the outer query has only 1 or 0 articles in the same industry with a later date, then it has to be among the latest two for that industry.