InstantForum.NET v2.9.0SQLServerCentralhttp://www.sqlservercentral.com/Forums/notifications@sqlservercentral.comMon, 20 May 2013 12:54:10 GMT20http://www.sqlservercentral.com/Forums/Topic404608-555-1.aspxtop 5th Salary of every departmentSELECT * FROM (SELECT EMP_ID, DEPT_ID, SALARY, NAME,RANK() OVER (PARTITION BY DEPT_ID ORDER BY SALARY DESC) AS RANKIDFROM EMP) A WHERE RANKID = 5Mon, 17 Aug 2009 23:49:50 GMTAmulya Duttahttp://www.sqlservercentral.com/Forums/Topic404608-555-1.aspxMay be the old thread but I want to show you other approaches[url]http://sqlblogcasts.com/blogs/madhivanan/archive/2007/08/27/find-nth-maximum-value.aspx[/url]Fri, 19 Sep 2008 06:59:48 GMTMadhivanan-208264http://www.sqlservercentral.com/Forums/Topic404608-555-1.aspxHi, This query is absolutely woking fine ...but if two employees r aving the same salary then it does not show correct result.. actuly it should show both employees for that rank. this can be found using this query...:--&gt;select * from @table where salary in (select top 1 salary from @table where salary in (select top [b][size="4"]3[/size][/b] salary from @Table order by salary [b]desc[/b] ) order by salary ) Looks complicated but not so ...Sun, 21 Oct 2007 23:25:18 GMTDhananjay Kulkarni-482726http://www.sqlservercentral.com/Forums/Topic404608-555-1.aspxComments posted to this topic are about the item [B]<A HREF="/scripts/Miscellaneous/31921/">Find Nth Maximum from a Table...</A>[/B]Sat, 29 Sep 2007 23:57:00 GMTHari.Sharma