How get relationship's startNodeId and endNodeId using Neo4jClient?

var query = client.Cypher
             .Match("(n1)-[r]-[n2]")
             .Where("n1.NodeId={startNodeId} and n2.NodeId={endNodeId}")
             .WithParam("startNodeId",startNodeId)
             .withParam("endNodeId",endNodeId)
             .return((r,n1,n2)=>new {Edge=r.As<Node<string>>(),Type=r.Type()});

By this way,I can only get the relationship's Label and Properties.But I also want to get the relationship's startNodeId and endNodeId. By the way,I can get the relationship's startNodeId and endNodeId by using REST API. How can anyone help me?

Z.Tom

Solution

It depends on what you're meaning by startNodeId - if you mean the Neo4j Ids, then you'll be wanting to use the Relationship<T> class:

client.Cypher
    .Match("(n1)-[r]-(n2)")
    //Your where stuff here.
    .Return(r => r.As<RelationshipInstance<string>());

The T should be a type you've put on the relationship, but if you haven't got that, just use object, you will need to use the Relationship<T> type, you can't use the non-generic version.

If the startNodeId is a property on a type, then you either do r => r.As<YourType>() or r => r.As<Relationship<YourType>>().

As a side note, you can use Neo4jClients ability to parse parameters and save some code (also make it more compile safe) with respect to your Where clauses.

I would write your code as:

var query = client.Cypher
    .Match("(n1)-[r]-[n2]")
    .Where((NodeType n1) => n1.NodeId == startNodeId)
    .AndWhere((NodeType n2) => n2.NodeId == endNodeId)
    .Return(r => new { Edge = r.As<RelationshipInstance<object>>(), Type = r.Type()});

which auto parses the parameters.