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Posted to dev@tinkerpop.apache.org by pieter gmail <pi...@gmail.com> on 2017/05/28 08:45:45 UTC
RE: hasId
Hi,
The following code illustrates my concern/confusion.
@Test
public void testHasId() {
final TinkerGraph graph = TinkerGraph.open();
Vertex a = graph.addVertex(T.label, "A");
Vertex b = graph.addVertex(T.label, "B");
List<Vertex> vertices =
graph.traversal().V().hasId(a.id()).hasId(b.id()).toList();
Assert.assertTrue(vertices.isEmpty());
}
The test fails as the both vertices are returned.
Is this expected, I expected 'and' not 'or' behavior.
Similar to,
@Test
public void testHasLabel() {
final TinkerGraph graph = TinkerGraph.open();
Vertex a = graph.addVertex(T.label, "A");
Vertex b = graph.addVertex(T.label, "B");
List<Vertex> vertices =
graph.traversal().V().hasLabel("A").hasLabel("B").toList();
Assert.assertTrue(vertices.isEmpty());
}
This one passes.
I checked the docs,
|hasLabel(labels...)|: Remove the traverser if its element does not have
any of the labels.
|hasId(ids...)|: Remove the traverser if its element does not have any
of the ids.
Seems they should behave the same?
I am working on version 3.2.4
Thanks
Pieter
Re: hasId
Posted by Daniel Kuppitz <me...@gremlin.guru>.
I created the ticket: https://issues.apache.org/jira/browse/TINKERPOP-1681
On Sun, May 28, 2017 at 11:12 AM, Daniel Kuppitz <me...@gremlin.guru> wrote:
> Hi Pieter,
>
> that's a nasty bug. I just verified that it's also a bug in master/:
>
> gremlin> graph = TinkerGraph.open()
> ==>tinkergraph[vertices:0 edges:0]
> gremlin> a = graph.addVertex(label, "A")
> ==>v[0]
> gremlin> b = graph.addVertex(label, "B")
> ==>v[1]
> gremlin> graph.traversal().V().hasId(a.id()).hasId(b.id())
> ==>v[0]
> ==>v[1]
> gremlin> graph.traversal().V().hasId(a.id()).hasId(b.id()).explain()
> ==>Traversal Explanation
> ============================================================
> ==============================
> Original Traversal [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
>
> ConnectiveStrategy [D] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> MatchPredicateStrategy [O] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> FilterRankingStrategy [O] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> InlineFilterStrategy [O] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> IncidentToAdjacentStrategy [O] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> AdjacentToIncidentStrategy [O] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> RepeatUnrollStrategy [O] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> CountStrategy [O] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> PathRetractionStrategy [O] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> LazyBarrierStrategy [O] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> TinkerGraphCountStrategy [P] [GraphStep(vertex,[]),
> HasStep([~id.eq(0), ~id.eq(1)])]
> TinkerGraphStepStrategy [P] [TinkerGraphStep(vertex,[0, 1])]
> ProfileStrategy [F] [TinkerGraphStep(vertex,[0, 1])]
> StandardVerificationStrategy [V] [TinkerGraphStep(vertex,[0, 1])]
>
> Final Traversal [TinkerGraphStep(vertex,[0, 1])]
> gremlin> Gremlin.version()
> ==>3.3.0-SNAPSHOT
>
> Can you create a ticket?
>
> Cheers,
> Daniel
>
>
> On Sun, May 28, 2017 at 10:45 AM, pieter gmail <pi...@gmail.com>
> wrote:
>
>> Hi,
>>
>> The following code illustrates my concern/confusion.
>>
>> @Test
>> public void testHasId() {
>> final TinkerGraph graph = TinkerGraph.open();
>> Vertex a = graph.addVertex(T.label, "A");
>> Vertex b = graph.addVertex(T.label, "B");
>>
>> List<Vertex> vertices = graph.traversal().V().hasId(a.id
>> ()).hasId(b.id()).toList();
>> Assert.assertTrue(vertices.isEmpty());
>> }
>>
>> The test fails as the both vertices are returned.
>> Is this expected, I expected 'and' not 'or' behavior.
>>
>> Similar to,
>>
>> @Test
>> public void testHasLabel() {
>> final TinkerGraph graph = TinkerGraph.open();
>> Vertex a = graph.addVertex(T.label, "A");
>> Vertex b = graph.addVertex(T.label, "B");
>>
>> List<Vertex> vertices = graph.traversal().V().hasLabel
>> ("A").hasLabel("B").toList();
>> Assert.assertTrue(vertices.isEmpty());
>> }
>>
>> This one passes.
>>
>> I checked the docs,
>>
>> |hasLabel(labels...)|: Remove the traverser if its element does not have
>> any of the labels.
>> |hasId(ids...)|: Remove the traverser if its element does not have any of
>> the ids.
>>
>> Seems they should behave the same?
>>
>> I am working on version 3.2.4
>>
>> Thanks
>> Pieter
>>
>>
>
Re: hasId
Posted by Daniel Kuppitz <me...@gremlin.guru>.
Hi Pieter,
that's a nasty bug. I just verified that it's also a bug in master/:
gremlin> graph = TinkerGraph.open()
==>tinkergraph[vertices:0 edges:0]
gremlin> a = graph.addVertex(label, "A")
==>v[0]
gremlin> b = graph.addVertex(label, "B")
==>v[1]
gremlin> graph.traversal().V().hasId(a.id()).hasId(b.id())
==>v[0]
==>v[1]
gremlin> graph.traversal().V().hasId(a.id()).hasId(b.id()).explain()
==>Traversal Explanation
==========================================================================================
Original Traversal [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
ConnectiveStrategy [D] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
MatchPredicateStrategy [O] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
FilterRankingStrategy [O] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
InlineFilterStrategy [O] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
IncidentToAdjacentStrategy [O] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
AdjacentToIncidentStrategy [O] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
RepeatUnrollStrategy [O] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
CountStrategy [O] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
PathRetractionStrategy [O] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
LazyBarrierStrategy [O] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
TinkerGraphCountStrategy [P] [GraphStep(vertex,[]),
HasStep([~id.eq(0), ~id.eq(1)])]
TinkerGraphStepStrategy [P] [TinkerGraphStep(vertex,[0, 1])]
ProfileStrategy [F] [TinkerGraphStep(vertex,[0, 1])]
StandardVerificationStrategy [V] [TinkerGraphStep(vertex,[0, 1])]
Final Traversal [TinkerGraphStep(vertex,[0, 1])]
gremlin> Gremlin.version()
==>3.3.0-SNAPSHOT
Can you create a ticket?
Cheers,
Daniel
On Sun, May 28, 2017 at 10:45 AM, pieter gmail <pi...@gmail.com>
wrote:
> Hi,
>
> The following code illustrates my concern/confusion.
>
> @Test
> public void testHasId() {
> final TinkerGraph graph = TinkerGraph.open();
> Vertex a = graph.addVertex(T.label, "A");
> Vertex b = graph.addVertex(T.label, "B");
>
> List<Vertex> vertices = graph.traversal().V().hasId(a.id()).hasId(
> b.id()).toList();
> Assert.assertTrue(vertices.isEmpty());
> }
>
> The test fails as the both vertices are returned.
> Is this expected, I expected 'and' not 'or' behavior.
>
> Similar to,
>
> @Test
> public void testHasLabel() {
> final TinkerGraph graph = TinkerGraph.open();
> Vertex a = graph.addVertex(T.label, "A");
> Vertex b = graph.addVertex(T.label, "B");
>
> List<Vertex> vertices = graph.traversal().V().hasLabel
> ("A").hasLabel("B").toList();
> Assert.assertTrue(vertices.isEmpty());
> }
>
> This one passes.
>
> I checked the docs,
>
> |hasLabel(labels...)|: Remove the traverser if its element does not have
> any of the labels.
> |hasId(ids...)|: Remove the traverser if its element does not have any of
> the ids.
>
> Seems they should behave the same?
>
> I am working on version 3.2.4
>
> Thanks
> Pieter
>
>