Forum: Ruby on Rails Some help with fun SQL stuff...

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75416846d25d74d7f8d3c68df06b435e?d=identicon&s=25 Pat M. (pergesu)
on 2006-06-13 01:56
(Received via mailing list)
I have one table that stores data related to math problems, and
another table that stores users' answers to those problems.  To make
this very simple, we'll say that the math table has two numbers that
get subtracted, and the answers table has a field that can either be
"pos" or "neg"

math
_____
2      5
8      3

Okay so if we do the math, we know that the correct answer to record 1
(2 -5) is "neg", and the correct answer to record 2 (8-3) is "pos".
What I'd like to do is get the user's answers to these questions, and
determine what % they get correct.  I really want to do this in as few
queries as possible though.

Right now I think the easiest way is to just change the table a bit so
it keeps the correct answer.  Then I can do two queries:

SELECT COUNT(*) FROM answers WHERE user_id=5;

SELECT COUNT(SELECT answer FROM answers, problems WHERE
answers.answer=problems.answer AND answers.user_id=5));

The first one gets the number of total questions answered, the second
one gets the number of ones correctly answered.  The first query - the
second query gives me incorrect answers.

Also I'm not entirely sure that the syntax for #2 is valid, I think it
is though.

First of all, is there a fast way that I can do this without having to
create another field representing the answer?  I'm kind of thinking
not...I'd be performing these queries a lot, makes sense just to keep
the answer in the db so I can index it, rather than calculating it
every time.  Secondly, can I do this with AR?  Don't know if it can
handle subqueries.

Pat
880f96b24ef64881d9ee2e9f062d59a8?d=identicon&s=25 Glenn C. (glenn)
on 2006-06-13 03:40
Pat Maddox wrote:
> I have one table that stores data related to math problems, and
> another table that stores users' answers to those problems.  To make
> this very simple, we'll say that the math table has two numbers that
> get subtracted, and the answers table has a field that can either be
> "pos" or "neg"
>
> math
> _____
> 2      5
> 8      3
>
> Okay so if we do the math, we know that the correct answer to record 1
> (2 -5) is "neg", and the correct answer to record 2 (8-3) is "pos".
> What I'd like to do is get the user's answers to these questions, and
> determine what % they get correct.  I really want to do this in as few
> queries as possible though.
>
> Right now I think the easiest way is to just change the table a bit so
> it keeps the correct answer.  Then I can do two queries:
>
> SELECT COUNT(*) FROM answers WHERE user_id=5;
>
> SELECT COUNT(SELECT answer FROM answers, problems WHERE
> answers.answer=problems.answer AND answers.user_id=5));
>
> The first one gets the number of total questions answered, the second
> one gets the number of ones correctly answered.  The first query - the
> second query gives me incorrect answers.
>
> Also I'm not entirely sure that the syntax for #2 is valid, I think it
> is though.
>
> First of all, is there a fast way that I can do this without having to
> create another field representing the answer?  I'm kind of thinking
> not...I'd be performing these queries a lot, makes sense just to keep
> the answer in the db so I can index it, rather than calculating it
> every time.  Secondly, can I do this with AR?  Don't know if it can
> handle subqueries.
>
> Pat

Sorry this is a little confusing particularly about having a (pos) or
(neg) field. Could you explain a little more?


If we keep the answer in the problem table then I would try something
like this.

correctly_answered = Answers.find_by_sql ( "select count(a.*) as
total_answers,
count(p.*) as correctly_answered " +
" from answers as a left outer join problems as p on * a.problem_id =
p.id  )" +
" where a.answer = p.answer " +
"   and a.user_id = ' + user_id )


if a computed answer is needed then

correctly_answered = Answers.find_by_sql ( "select count(a.*) as
total_answers,
count(p.*) as correctly_answered " +
" from answers as a left outer join problems as p on * a.problem_id =
p.id  )" +
" where (a.number_to_subtract_from -
a.number_for_the_amount_to_subtract) = p.answer " +
"   and a.user_id = ' + user_id )
75416846d25d74d7f8d3c68df06b435e?d=identicon&s=25 Pat M. (pergesu)
on 2006-06-13 04:28
(Received via mailing list)
On 6/12/06, Glenn Cadman <glenn.cadman@gmail.com> wrote:
> > 8      3
> > SELECT COUNT(*) FROM answers WHERE user_id=5;
> >
> (neg) field. Could you explain a little more?
Okay let's say my app shows (4 - 2) on the page.  The users of the app
don't have to figure out the answer, but simply tell me if the result
is positive or negative.

Pat
880f96b24ef64881d9ee2e9f062d59a8?d=identicon&s=25 Glenn C. (glenn)
on 2006-06-13 05:05
Pat Maddox wrote:
> On 6/12/06, Glenn Cadman <glenn.cadman@gmail.com> wrote:
>> > 8      3
>> > SELECT COUNT(*) FROM answers WHERE user_id=5;
>> >
>> (neg) field. Could you explain a little more?
> Okay let's say my app shows (4 - 2) on the page.  The users of the app
> don't have to figure out the answer, but simply tell me if the result
> is positive or negative.
>
> Pat


Assuming you are talking about kids students then in postgresql (i have
no idea about mysql) then you could try the "sign" function

sign(question.from_number - question.subtract_by) =
sign(answer.answer_value)

Where answer.answer_value is 1,-1 (is zero possible?)

check_box("answer", "answer_value", { }, "-1", "1" )
880f96b24ef64881d9ee2e9f062d59a8?d=identicon&s=25 Glenn C. (glenn)
on 2006-06-13 05:07

correctly_answered = Answers.find_by_sql ( "select count(a.*) as
total_answers,
count(p.*) as correctly_answered " +
" from answers as a left outer join problems as p on * a.problem_id =
p.id  )" +
" where sign(a.number_to_subtract_from -
a.number_for_the_amount_to_subtract) = sign(p.answer) " +
"   and a.user_id = ' + user_id )
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