Re: tricky SQL

From: bcjm <oldcar_at_hotmail.com>
Date: 4 Mar 2005 07:05:56 -0800
Message-ID: <1109948756.491657.115600@o13g2000cwo.googlegroups.com>

Thanks. I figure out the answer.

select p,

max(decode(rn, 1, c)) test1,
max(decode(rn, 1, g)) grade1,
max(decode(rn, 2, c)) test2,
max(decode(rn, 2, g)) grade2,
max(decode(rn, 3, c)) test3,
max(decode(rn, 3, g)) grade3
from (select p, c, g, row_number() over (partition by p order by c) rn
       from t1)

 group by p Received on Fri Mar 04 2005 - 09:05:56 CST