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    <p>It seems to me that the question calls for a query that would return best, and second best in the same row for each month and year, like so:</p> <pre><code>month, year, best, second best ... ... </code></pre> <p>and not two rows for the same month and year containing best and second best value.</p> <p>This is the solution that I came up with, so if anyone has a simpler way of achieving this, I would like to know.</p> <pre><code>with ranks as ( select year(entrydate) as [year], month(entrydate) as [month], views, rank() over (partition by year(entrydate), month(entrydate) order by views desc) as [rank] from product ) select t1.year, t1.month, t1.views as [best], t2.views as [second best] from ranks t1 inner join ranks t2 on t1.year = t2.year and t1.month = t2.month and t1.rank = 1 and t2.rank = 2 </code></pre> <p>EDIT: Just out of curiosity I did a bit more testing and ended up with a simpler variation on the <a href="https://stackoverflow.com/questions/4440598/query-to-find-the-first-and-second-largest-value-from-a-group/4443936#4443936">Stephanie Page's answer</a> that doesn't use an aditional subquery. And I changed the rank() function to row_number() as it doesn't work when two max values are the same.</p> <pre><code>with ranks as ( select year(entrydate) as [year], month(entrydate) as [month], views, row_number() over (partition by year(entrydate), month(entrydate) order by views desc) as [rank] from product ) select t1.year, t1.month, max(case when t1.rank = 1 then t1.views else 0 end) as [best], max(case when t1.rank = 2 then t1.views else 0 end) as [second best] from ranks t1 where t1.rank in (1,2) group by t1.year, t1.month </code></pre>
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    1. COYes, this is definitely a working solution, thx. I think there are several similar solutions using joins, but i am still wondering (or better said, just curious) if there is one without them?
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    2. COI came accross similar problems over the years and was also wondering if there is a simpler solution (without joins maybe), but couldn't fine one. And as it turns out the join solution, although it may seem like a kludge, work perfectly and fast enough.
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    3. CO@Helena: btw vote and/or accept the answer if you think it is useful, and welcome to stackoverflow
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