Got it. Thanks!
Rick >> -1
A
On Thu, Nov 17, 2011 at 6:36 PM, Marko Rodriguez <okramma...@gmail.com> wrote:
>> What exactly does iterate() do?
>
> A Gremlin expression yields a Pipeline and
> Pipeline<S,E> implements Iterator<S,E>
>
> Thus, a Gremlin expression doesn't return the results of a traversal, only a
> lazy iterator for .next()'ing results. For people doing a traversal to yield
> a sideEffect (e.g. aggregate, groupCount, etc.), iterate() is there to
> "while(hasNext()) { next() }" for you to "fill up" your side effect data
> structure.
>
> Finally, the Gremlin REPL automagically iterates and System.out.printlns()
> any objects in an Iterator it gets so thats why the behavior in the Gremlin
> REPL looks like a Gremlin expression returns results.
>
> Hope that is clear,
> Marko.
>
> http://markorodriguez.com
>
>> A
>>
>> On Thu, Nov 17, 2011 at 4:07 PM, Marko Rodriguez <okramma...@gmail.com>
>> wrote:
>>> Hey Alfredas,
>>>
>>> Be sure to iterate your pipeline....
>>>
>>> x = []; g.v(1).out("from").out("to").aggregate(x).loop(3){it.loops <
>>> 4}.iterate(); x
>>>
>>> * NOTE: You can also do:
>>> g.v(1).out("from").out("to").aggregate(x).loop(3){it.loops < 4} >>-1
>>> but the >> convention is no longer with us in Gremlin 1.4-SNAPSHOT.
>>>
>>> HTH,
>>> Marko.
>>>
>>> http://markorodriguez.com
>>>
>>> On Nov 17, 2011, at 7:21 AM, Alfredas Chmieliauskas wrote:
>>>
>>>> Dear all,
>>>> This concerns gremlin plugin and the script engine.
>>>> Maybe there's an explanation for this behavior:
>>>> 1) gremlin> x = [];
>>>> g.v(1).out("from").out("to").aggregate(x).loop(3){it.loops < 4};
>>>> gremlin> x;
>>>>
>>>> ==> v[7]
>>>> ==> v[3]
>>>> ==> v[5]
>>>>
>>>> 2) gremlin> x = [];
>>>> g.v(1).out("from").out("to").aggregate(x).loop(3){it.loops < 4}; x;
>>>> returns nothing...
>>>>
>>>> Thanks,
>>>>
>>>> Alfredas
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