final Graph newGraph = TinkerGraph.open();
newGraph.io(IoCore.graphml()).readGraph("example.xml");
And got the next exception on Edge:
Exception in thread "main" java.lang.IllegalArgumentException: Label can not be null
at org.apache.tinkerpop.gremlin.structure.Element$Exceptions.labelCanNotBeNull(Element.java:137)
at org.apache.tinkerpop.gremlin.structure.util.ElementHelper.validateLabel(ElementHelper.java:62)
at org.apache.tinkerpop.gremlin.tinkergraph.structure.TinkerHelper.addEdge(TinkerHelper.java:51)
at org.apache.tinkerpop.gremlin.tinkergraph.structure.TinkerVertex.addEdge(TinkerVertex.java:125)
at org.apache.tinkerpop.gremlin.structure.io.graphml.GraphMLReader.readGraph(GraphMLReader.java:190)
at org.apache.tinkerpop.gremlin.structure.io.graphml.GraphMLIo.readGraph(GraphMLIo.java:86)
at mycompany.main(App2.java:35)
If I add the label to the edge, everything works.
My question is if there is any graphml or gremlin standard that requires to have a label. What is the motivation of this constrain?
Thank you!
On Jun 6, 2016, at 1:48 PM, Serge Vilvovsky <serg...@gmail.com> wrote:
If the standard does not require, is it any way to provide the default edge label for tinkerpop3 to avoid the exception?
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final Graph newGraph = TinkerGraph.open();
newGraph.io(IoCore.graphml()).readGraph("example.xml");
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Is it any way that I can add the default label in parser? Any examples of such parser configuration?
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final GraphTraversalSource g = newGraph.traversal();
GraphTraversal<Vertex, Path> result1 = g.V().as("v1").has("type","type1").out().as("v2").has("type", "type2").path();
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