CODEWITHSUNDEEP
groovygremlin
Satyam
Satyam

Reputation: 21

On gremlin console, why predicate 'eq' returning false inside 'is' while direct comparison returns true?

On gremlin version 3.4.0 with step is, the predicates are not working as expected.

if I do the following it is returning me true

gremlin>3.is(3)
==>true
gremlin>[3].getAt(0).is(3)
==>true

While the same comparisons with using predicates are not working and returns false

gremlin>3.is(eq(3))
==>false
gremlin>[3].getAt(0).is(eq(3))
==>false

Precisely I want to check the length of the node property value and validate based on a max length. 

gremlin>g.V(0).values('name').next().length().is(lte(20))
==>false

The above code always return false is the name is equal to test_name (with length 9). While if I do direct comparison (like shown below) it returns true

gremlin>g.V(0).values('name').next().length().is(9)
==>true

What am I doing wrong here?

Upvotes: 0

Views: 198

Answers (1)

stephen mallette
stephen mallette

Reputation: 46216

You are mixing Groovy and Gremlin. When you do this:

gremlin>3.is(3)
==>true
gremlin>[3].getAt(0).is(3)
==>true

you are not using Gremlin and that is therefore not the is() step. In that case you are doing a reference equality check using Groovy's is() method. That further explains your results as you go further down in your question as in:

gremlin>g.V(0).values('name').next().length().is(lte(20))
==>false

because as soon as you do next() you are no longer doing Gremlin. Nothing following that represent Gremlin steps, it's just Groovy code. You are calling the String.length() method and then the Groovy is() operator and comparing the int returned from length() to lte(20) which is an instance of P thus:

gremlin> "xyz".length().is(lte(20))
==>false

If you want to use Gremlin to check the length of a string then I'm not sure that there is a way to do so outside of using a lambda:

gremlin> g = TinkerFactory.createModern().traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
gremlin> g.V().values('name').filter{it.get().length()<=4}
==>lop
==>josh

Upvotes: 1

Related Questions