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Perform calulation and select items when condition is met

Posted on 2013-06-11
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Last Modified: 2013-06-11
I am using PL/SQL to write a query to extract the following:

I need to calculate:
round(t.fee_net/t.svngs_net,2)
then I need to select only those items where the result is = '.20'
I don't want to select any items where svngs_net is '0'
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Question by:AlphaMig1
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Accepted Solution

by:
johnsone earned 500 total points
ID: 39237722
select ...
from your_table
where round(t.fee_net/t.svngs_net,2) = .2
and svngs_net != 0
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LVL 13

Expert Comment

by:jonnidip
ID: 39237731
While I think there could be a performance issue (it should be checked by looking at the actual indexes of your tabled), I would do something like this:
select * from myTable where round(t.fee_net/t.svngs_net,2) = 0.20

Open in new window

Isn't it?

Regards.
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Author Comment

by:AlphaMig1
ID: 39237797
Thanks for your response jonnidip,

Using the formula you suggested, I get an error that divisor is '0' so I tried:

select
(case
      when b.svngs_net <> '0'
        then round(t.fee_net/t.svngs_net,2) else 0 end)as fee_percent

from myTable t

where
round(t.fee_net/t.svngs_net,2)='.20'

But is this the most efficient way to do this?
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LVL 13

Expert Comment

by:jonnidip
ID: 39237835
OK, I understand what was the problem with '0'.
You may go with your way of using a subquery, but you need to modify it as:
select 
		t.*,
		(case when b.svngs_net <> 0
				then round(t.fee_net/t.svngs_net,2) else 0 end)as fee_percent
from myTable t
where 
t.fee_percent=0.20

Open in new window


You have surrounded '0' and '.20' with single quotes: are these fields varchar or numeric?

Regards.
0
 

Author Comment

by:AlphaMig1
ID: 39237886
Yes, they are numeric.  And your suggestion worked!
Thank you
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