David,
Adding to what Michael said, an idiomatic pattern I've been using is:
obj = kv.bucket(b).get(name)
data = f.get_data() or {'my_list': []}
data['my_list'].append(something)
obj.set_data(data)
obj.store()
[usually it's `f.get_data() or {}` when I use it, but you should see the
pattern.]
Hope that helps save a couple lines at a time,
--
Adam Lindsay
On Saturday, 29 September 2012 at 03:23, David Lowell wrote:
> Hi folks,
>
> Simple question here. Based on the python client documentation, I'm unclear
> on the canonical means to perform in python a read-modify-write cycle on an
> object stored in Riak.
>
> What seems intuitive to me:
>
> obj = kv.bucket(b).get(name)
> if f.exists():
> data = f.get_data()
> # todo: check for siblings, and resolve
> else:
> data = {'my_list': []}
> data['my_list'].append(something)
>
> obj.set_data(data)
> obj.store()
>
> My assumption here is that the RiakObject 'obj' has internally kept the
> vector clock for this key, and when this update gets stored, it will do the
> right thing. However, the documentation doesn't say that's what will happen.
> In fact, I can't find anything in the documentation that makes it clear how
> to do this read-modify-write.
>
> There is this piece of documentation for the python client API:
>
> | populate(self, Result)
> | Populate the object based on the return from get.
> |
> | If None returned, then object is not found
> | If a tuple of vclock, contents then one or more
> | whole revisions of the key were found
> | If a list of vtags is returned there are multiple
> | sibling that need to be retrieved with get.
>
> This is obviously not very clear. This sounds like it has something to do
> with read-modify-write, but who knows.
>
> So, is my intuition in the code above correct? Or should we write back a
> modified object through some other means like populate()?
>
> Thanks for the guidance,
>
> Dave
>
> --
> Dave Lowell
> [email protected] (mailto:[email protected])
>
>
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>
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