Ordering via val variable doesn't seem to work

Users Dgraph
1

Schema:

    name: string @index(hash).
    nickname: string @index(hash).
    age: string .
    class: string @index(exact, trigram) .
    type TypeName {
        name: string
        class: string
        nickname: string
        age: int
    }

Data:

        <1> <name> "Anurag" .
        <1> <class> "first" .
        <1> <nickname> "Anu" .
        <1> <age> "10" .
        <1> <dgraph.type> "TypeName" .

        <2> <name> "Brad" .
        <2> <class> "second" .
        <2> <nickname> "BD" .
        <2> <age> "20" .
        <2> <dgraph.type> "TypeName" .

Query:

    q1(func: has(name)) {
        v as name
        a as age
    }
    q2(func: has(name), orderdesc: val(v)) { //Works if I pass orderdesc: name
        name
        age
        val(v)
        val(a)
    }

Output:

{
  "q1": [
    {
      "age": "10", 
      "name": "Anurag"
    }, 
    {
      "age": "20", 
      "name": "Brad"
    }
  ], 
  "q2": []
}

Is there something I am missing, I don’t know why q2 doesn’t seem to work?

2

I feel this is a bug and not a feature we don’t support because I did see my matrix getting correctly ordered uptil here

3

You have to do it via multiple blocks
e.g

{    
  G as q1(func: has(name)) {
        v as name
        a as age
    }
    q2(func: uid(G), orderdesc: val(v)) {
        name
        age
        val(v)
        val(a)
    }
}

Without the G reference (that works as a “map”) it won’t apply anything in the query block. The variables are based on levels and a kind of uid mapping.

This avoids things like

{    
   q1(func: has(name)) {
        v as name
        a as age
    }
    q2(func: has(title), orderdesc: val(v)) { 
        name
        age
        val(v)
        val(a)
    }
}

This query above would break anyway. Cuz it doesn’t has references.

4

It would work if I modify my query to be:

    f as q1(func: has(name)) {
        v as name
        a as age
    }
    q2(func: uid(f), orderdesc: val(v)) { //Works if I pass orderdesc: name
        name
        age
        val(v)
        val(a)
    }

Weird…

5

You mean the above query would break as well because it doesn’t have a map between two queries. Also, we do not want to order results of one query via variables defined in an entirely different query.

Is that correct understanding?

6

Yes, and potentially, the second block (using has) would bring more other new nodes. That would break the query. The right way is by “Mapping” it.

yep

2 Likes