My mistake not making it clear enough.

Actually they have some attributes.

Let's say, 

A has a simple attribute called A.name and C also has a attribute C.name.

Then A has_many C, which means C.a_id exists.

In my case, my requirement is that if users enter C2.name、 C3.name and 
C4.name, I should show them A1.name or whatever attributes A might have 
that needs to be displayed.

Anyway, thanks for your reply.

On Wednesday, January 13, 2016 at 4:47:35 PM UTC+8, Colin Law wrote:
>
> On 13 January 2016 at 07:22, Lei Zhang <ray.zh...@gmail.com <javascript:>> 
> wrote: 
> > I am going to implement a small and simple search feature in my project. 
> But 
> > got a question about the scenario I have right now. So hope to get some 
> tips 
> > from here. 
> > 
> > First of all, this is a rails app. 
> > 
> > There are two models, A and C. 
> > 
> > A has_many C. 
> > 
> > That's to say, it might have some situation like this: 
> > 
> > A1 related to C1, C2, C3, C4 
> > A2 related to C2, C3, C4, C5 
> > A3 related to C2, C3, C4, C8 
> > A4 related to C3, C4, C8, C9 
> > A5 related to C4, C8, C9, C20 
>
> For that you need A has_and_belongs_many C, or (my preferred solution) 
> introduce a join model/table and use has_many through.  You can't use 
> has_many as that would require 
> C belongs_to A 
> which is not correct. 
>
> > 
> > Use C2, C3, C4 as keyword, I can get A1, A2, A3 
> > Use C4, C8, C9 as keyword, I can get A4, A5 
> > 
> > I consider that this can be done by using something straightforward like 
> > A.Cs.includes?(C2,C3,C4). 
> > 
> > But that requires an iteration of A. And I concern that it would have 
> some 
> > performance issue. 
> > 
> > Any advice or idea about this requirement ? 
>
> I don't understand exactly what your requirement is.  in particular 
> what do you mean by "use C2, C3, C4 as keyword", when each of the Cs 
> is a record in the db. 
> If you are trying to get all the As related to records C2, C3 and C4 
> then that is something like 
> (c2.as + c3.as +c4.as).uniq 
> though that may not be the most efficient way. 
>
> Colin 
>

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