How to traverse without indexed value? - node.js

Question

I have a scenario that I want to query for getting all the vertices which are not having a specific property key and value.So, what I have tried is
g.V().hasLabel('Persona').hasNot('isTempDob').count()
resulted in
Could not find a suitable index to answer graph query and graph scans are disabled: [(~label = Persona)]:VERTEX
So I have tried for the simpler one by following the documentation like getting the count of vertices with label name Persona. The query is
g.V('Persona').count()
resulted in
0
I don't know why I am not able to traverse without any indexed value? Any help would be greatly appreciated.Thanks in advance.

Answer 1

Perhaps it is just a mis-type but g.V('Persona').count() does not search labels - it looks for vertices with a unique identifier of "Persona". As a result I suspect that the answer of "0" was correct.
I don't know why I am not able to traverse without any indexed value?
JanusGraph (and other graph systems) prevent traversals without indexed values as a guardrail to prevent users from accidentally executing long run queries. If you have a small graph with just thousands vertices then it's likely you find this feature an annoyance, but when you have billions of vertices it can be a great helper as such a query as the one you've written there would take hours and hours to complete and would not be the recommended approach to get that answer. If you are doing a global scan of all the vertices in the graph then you should be executing your Gremlin over Spark to enjoy it's parallel processing capabilities, turning those hours and hours into minutes.
Going back to the case of a small graph, you can temporarily disable the guardrail by configuring JanusGraph with query.force-index=false to allow the query to execute. That should allow your query to work without that error/warning.

Answer 2

I tested in the gremlin console docker.
The identifier is a unique ID for each vertex. When creating a vertex, you can either specify an ID or generate one automatically. For example,
gremlin> saturn = g.addV("character").property(T.id, 1).property('name', 'saturn').property('age', 10000).property('type', 'titan').next();
==>v[1]
If you create a vertex in the above way, you'll get v[1] as the ID cause you specify it.
The following didn't specify ID, then gremlin will generate one for you:
gremlin> g.addV("character").property('name', 'prometheus').property('age', 1000).property('type', 'god').next()
==>v[0]
Assume it's first vertex, then its ID will be 0.
In your scenario, if you are querying all vertices have the same tag and exclude a specified name, you can try this(I'm using the JanusGraph toy graph gods):
gremlin> g.V().has('character','name',neq('hercules')).valueMap();
==>[name:[prometheus],type:[god],age:[1000]]
==>[name:[saturn],type:[titan],age:[10000]]
==>[name:[jupiter],type:[god],age:[5000]]
==>[name:[neptune],type:[god],age:[4500]]
==>[name:[alcmene],type:[human],age:[45]]
==>[name:[pluto],type:[god],age:[4000]]
==>[name:[nemean],type:[monster],age:[20]]
==>[name:[hydra],type:[monster],age:[0]]
==>[name:[cerberus],type:[monster],age:[0]]
The above query finds all vertices tagged character and exclude name hercules. Or you can use the following query to do the same thing.
gremlin> g.V().hasLabel('character').has('name',neq('hercules')).valueMap();
The reason your query didn't work is that hasNot(key) must be used with key rather than values.
Let me know if it helps.