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プログラマーと熱狂的なプログラマーのためのサイトに関する質問と回答

SELECT * FROM
(
SELECT TOP 100 
   'Silver' AS [Type],
   UserId AS [User Link],
   Count(*) as [Count]
FROM Badges b
INNER JOIN users u ON u.id = b.userid
      WHERE location like '%##Location##%' AND Class=2
GROUP BY UserId
ORDER BY [Count] DESC
) s

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