Message-ID: <1608.219.65.253.230.1050435668.squirrel@mail.trade-india.com>
Date: Wed, 16 Apr 2003 01:11:08 +0530 (IST)
Subject: Re: Percentage of Total Occurances
From: <mallah@trade-india.com>
To: <adam@tritus.ca>
In-Reply-To: <30270000.1050433172@saturn>
References: <6790000.1050418432@saturn>
	<1177.219.65.253.230.1050426637.squirrel@mail.trade-india.com>
	<30270000.1050433172@saturn>
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well if everything in a single query is that imp u may do this subselect:

 select answer_id ,  1.0*count(*) / (select count(*) from polls where poll_=
id =3D ?)
      from polls where poll_id =3D ? group by answer_id ;

and if u need to trim the trailing digits after decimal
do to_char(value , '99.99')

since u have not provided any sample data i cann't test it.


> On 04/15/03 22:40:37 +0530 mallah@trade-india.com wrote:
>
>>> I have a table being used for poll results:
>>>
>>> id
>>> poll_id
>>> question_id
>>> answer_id
>>> datestamp
>>>
>>> I what to figure out the percentage or respondants that gave each answe=
r.  This is based on
>>> occurance rather than values.
>>
>>
     select answer_id ,  1.0*count(*) / (select count(*) from polls where p=
oll_id =3D ?)
      from polls where poll_id =3D ? group by answer_id ;
>>
>> gives number of response of each answer
>>
>> total_response =3D select count(*) from polls where poll_id =3D ?
>>
>> getting %age shud be trivial.
>>
>> Dunno if i missed your question .
>
> I'm looking to do everything in a single query.
>
> And get something like:
>
> answer_id -> percentage
>
> Thanks,
>
> A.
>
> --
> Adam Sherman
> Tritus CG Inc.
> http://www.tritus.ca/
> +1 (613) 797-6819
>
>
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