Depends on the shape of your data. I recommend you to use indexed key or
relation index "pattern", only if the size is bigger than 100.
IF (List<Key<B>>.size() > 100) USE indexed property or relation index.
I hope it will help.
2012/2/9 Robert Kluin <robert.kl...@gmail.com>
> Hi Steven,
> Some responses inline.
>
> Robert
>
> On Wed, Feb 8, 2012 at 11:26, steven <jobshif...@gmail.com> wrote:
> > Hi:
> >
> > Thank you in advance
> >
> > I have a question about Datastore API calls.
> >
> > Suppose I have a one-to-many relation ship, with two solutions as
> > below:
> >
> > 1) solution 1
> > class A
> > {
> > set<key> keys_for_B
> > }
> >
> > the instance is a1 contains <b1, b2, b3>
> >
> > if I query for a1, and then I can get keys of b1, b2, b3; and then I
> > read the b1, b2, b3.
> > Does it mean there are 4 calls to datastore API calls?
>
> If you *query* for a1, then do a bulk get on those keys you'll have:
> query -> 1 scan + 1 fetch = 2 reads
> bulk get -> 3 reads
>
> So you'll have 5 reads. If you know A1's key, then you could do a get
> there too which would save 1 read.
>
>
> >
> >
> > 2) solution 2
> > I build a "big-table" as
> > a1, b1
> > a1, b2
> > a1, b3
> >
> > and then query as "select * from mytable where a_id = a1_id", to get 3
> > rows
> > I guess there is just one call, to datastore.
>
>
> Generally, I would prefer the first design over introducing a third
> lookup table.
>
>
> Another solution here might be to include a1 as a property on your "B"
> entities, possibly as an indexed list property. Then you could do one
> query for Bs where a = 'a1'. This would give you 4 reads, 1 for the
> scan + 3 for the entities.
>
>
>
>
> >
> > comparing solution 1 and 2, does it mean Google will charge me 4 times
> > money for solution 1?
> >
> > thank you very much,
> > Sincerely,
> > Steven
> >
> >
> >
> >
> >
> >
> >
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