following source
name gender
ramya female
ram male
deesha female
david male
kumar male
i want the target
male female
ram ramya
david deesha
kumar
any body give solution above question?
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Answer / annu
select decode(gender,'female',name) FEMALE,
decode(gender,'male',name) MALE from src;
| Is This Answer Correct ? | 1 Yes | 0 No |
use router transformation, In that 2 groups
1 is male, 2 is female,
target is single, but we drag the table twice
In that router 1st group condition
select * from table where gender='male'
2nd group condition
select * from table where gender='female'
| Is This Answer Correct ? | 2 Yes | 2 No |
Answer / moorthy g
Source-SQ-Exp-Rtr-Tgt(twice)
Exp:
Flag: DECODE(True, gender='MALE', 'Y','N')
RTR:
Make It Two Group
Flag value is Y then Male is one group.
Defaull is Female
Tgt:
Make It Two;
male - Tgt1
female - Tgt2
=================
| Is This Answer Correct ? | 2 Yes | 2 No |
Answer / jaspreet banga
after source qualifier, Use an expression transformer , in
that take an variable port name 'FLAG', and in that use the
decode function as 'FLAG = DECODE(GENDER=MALE,'M','F')'.
after that place an router transformer and for each flag
condition use seperate instance of target and hence you
will get what you want.
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / krishna
select * from (select distinct name male from fm where
gender='male'),(select distinct name female from fm e
where gender='female');
select decode(gender,'female',name)female,decode
(gender,'male',name)male from fm ;
| Is This Answer Correct ? | 3 Yes | 4 No |
Answer / murali udayagiri
You can use the below query in SQ trans if the source is
relational, if not u can do it in expr trans.
select coalesce((case when gender='male' then ename
end),',')||','||
coalesce((case when gender='female' then ename
end),'###') "male,female" from temp_emp1;
Thanks,
Murali Udayagiri
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / anwar
select name,
decode(gender,'male','MALE',
'female','FEMALE')GENDER from sOURCE;
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / star
Not sure if this has to be solved by SQL or by informatica -
and source is a table or file.
select * from (select lead(gender,0,0) over (partition by
gender), lead(gender,1,0) over (partition by gender) from
table1) where rownum=1
union all
select t1.name, t2.name from
(select name,row_number() over (partition by name) from
table1 where gender='male') t1,
(select name,row_number() over (partition by name) from
table1 where gender='female') t2
where
t1.name(+) = t2.name(+)
| Is This Answer Correct ? | 0 Yes | 2 No |
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