Answer Posted / kevin sharpless

-- This query wiil return any number of the n items by
using the max/decode statements

select max(decode(rnum,1,salary,null)) max1
,max(decode(rnum,2,salary,null)) max2
from (
select 1 grp_col
,salary
,row_number() over (order by salary asc) rnum
from (
select distinct salary
from emp
)
)
group by grp_col

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