Answer / lakshmi srinivas

source->copy->2 aggregators->join->target

1 aggregator->eno,max(sal),min(sal)

2 aggregator->eno,dno,max(sal)

by using max(sal) key, we can join both o/p of
aggregators,we can get that output...

Answer / shar

the question framed wrong
it should have been
empno,dept,sal
1,10,9
1,10,8
1,10,7
2,20,7
2,20,8.5
2,20,9
3,30,4
3,30,6
then expecting an ans is correct
o/p
empno,max(sal),min(sal),deptno
1,9,7,10
2,9,7,20
3,6,4,30
---xx----

where as in the o/p of user's question
we have
empno,max(sal),min(sal),deptno
1,9,3.5,20
2,8,4.5,20
3,6,4,10
here empno=1 his max(sal)=9 from deptno 20 and min(sal)=3.5 then its deptno=? actually we r giving a correct information but client will get confused max(sal) and min(sal) r both from 20 or different departments.
-----xxx----
even though client is expecting the same output then laxmi answer is correct

Answer / vinod upputuri

hai first use the copy stage. from that take three links.

in that first and second links are connected to aggregator
stages.

agg1-- max and min for sal group by empno...

agg2--max only group by empno...

then use lookup for agg2 and third link of copy stage...

in lookup join max{sal} to sal and get deptno...

finally, o/p links from agg1 and lookup are joined by
lookup.. join by max{sal}....


then u can get the desired o/p...

Answer / eswararao

create one group using column generator and then Using Aggregator stage select max sal,min sal

Answer / pankaj das, orator

Just take...
source -> aggregate stage -> target

And then in aggregate stage in output tab just specify

col name groupby derivation
------- ------- ----------
id group by id
max_sal max(sal)
min_sal min(sal)
deptno deptno

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