@Eric
By threshold, all I mean is the count of the documents returned and I am
not going to play with score. So if I have to commit my code to svn, whats
the best way to go about it? I know I have to discuss my design here which
would take atleast a couple of days. But is there special instructions that
I need to follow in order to stay in a direction from where I could commit
my code?


@John
Yes, thats definitely a solution but then I dont want to make two different
http requests. I want to make 1 request and all that I mentioned has to
happen.



On Mon, Apr 2, 2012 at 7:28 PM, Erick Erickson <[email protected]>wrote:

> Part of it depends on what you mean by "threshold". If it's
> just the number of matches, then fine. But if you're talking score
> here, be very, very careful. Scores are not an absolute measure
> of anything, they only tell you that "for _this_ query, the docs
> should be order this way".
>
> So I'd advise against any "query chain" based on scores
> as the threshold, if that's what you mean by "threshold".
>
> Best
> Erick
>
> On Mon, Apr 2, 2012 at 10:28 AM, Karthick Duraisamy Soundararaj
> <[email protected]> wrote:
> > Hi all,
> >        I am finding a need to merge the results of multiple queries to
> > accomplish a functionality similar to this :
> >
> >                     1. Make query 1
> >                     2. If results returned by query1 is less than a
> > certain threshold, then Make query 2
> >
> > Extending this idea, I want to be able to create a query chain, i.e,
> > provide a functionality where you could specify n queries and n-1
> > thresholds in a single url. Start querying in the order from 1 to n until
> > one of them produces results that exceed the threshold.
> >
> > PS: These n queries and n threshold are passed on a single url and each
> of
> > them could use different request handlers and therefore take a different
> > set of parameters.
> >
> > Any suggestions/thoughts/pointers as where to begin looking for will be
> of
> > great help!
> >
> > Thanks,
> > Karthick
>

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