4

Please login or register to vote for this query.

(click on this box to dismiss)

Select top N users by upvotes they received for their answers for a given tag located in the specified city or country

Q&A for linguists, teachers and learners wanting to discuss the finer points of the Portuguese language

-- Top N users for tag located in city / country
-- Created by Mihai Todor (http://stackoverflow.com/users/1174378/mihai-todor)

/*
Selects users based in the given city / country,
  who have the most upvoted answers for the given tag.

Usage:
Tag - tag name (mandatory)
Country, City - specify at least one
Results - number of desired results
*/

DECLARE @Tag NVARCHAR(35) = ##Tag:string##;
DECLARE @City NVARCHAR(100) = LTRIM(RTRIM(##City:string? ##))
DECLARE @Country NVARCHAR(100) = LTRIM(RTRIM(##Country:string? ##));
DECLARE @Results INT = ##Results?50##;

SELECT TOP (@Results)
  ROW_NUMBER() OVER(ORDER BY SUM(a.Score) DESC) AS [#],
  u.Id AS [User Link],
  SUM(a.Score) AS [Tag Reputation],
  u.Reputation AS [Overall Reputation]
FROM Posts AS p
INNER JOIN Posts AS a ON a.ParentId = p.id -- answers only
INNER JOIN PostTags AS pt ON p.Id = pt.PostId
INNER JOIN Tags AS t ON pt.TagId = t.Id
INNER JOIN Users AS u ON u.Id = a.OwnerUserId
  -- OR u.Id = a.LastEditorUserId -- Might be useful to play with
WHERE a.PostTypeId = 2 -- answers only (probably not needed; see join clause)
-- select all possible aliases for the given tag
AND t.TagName IN (
  SELECT TagName FROM Tags WHERE LOWER(TagName) LIKE LOWER('%' + @Tag + '%')
  UNION
  SELECT TargetTagName FROM TagSynonyms WHERE LOWER(SourceTagName) LIKE LOWER('%' + @Tag + '%')
)
AND (
  @City <> '' AND (LOWER(u.Location) LIKE LOWER('%' + @City + '%') OR
                   LOWER(u.AboutMe) LIKE LOWER('%' + @City + '%'))
  OR
  @Country <> '' AND (LOWER(u.Location) LIKE LOWER('%' + @Country + '%') OR
                      LOWER(u.AboutMe) LIKE LOWER('%' + @Country + '%'))
  OR
  @City = @Country -- don't care about location (blank input)
)
GROUP BY u.Id, u.Reputation
ORDER BY SUM(a.Score) DESC

fork query

edited apr 11 17

created aug 2 16 Mihai.Todor