Using Distinct in a standarized query

Neo4j Graph Platform
1

I have a program that expects a very simple output from Neo4j in the form of m,d,p

so this statement works fine.
MATCH (p:ProgNode {compileunit:'Way-Duy_User-Task-Manager', release:2})-[d:RUNS]->(m) RETURN m, d, p

But I have a situation where I need to display that information for each unique occurance of compileunit. I am trying to build one that uses a distinct directive (e.g. distinct (p.compileunit)) but I cannot figure out how to embed it so I maintain the output (m,d,p) the program is expecting.

Thanks

2

First you have to move the p element as your first grouping key, like this:

MATCH (p:ProgNode {compileunit:'Way-Duy_User-Task-Manager', release:2})-[d:RUNS]->(m) RETURN p.compileunit,d,m

After you have to put the operator DISTINCT like this

MATCH (p:ProgNode {compileunit:'Way-Duy_User-Task-Manager', release:2})-[d:RUNS]->(m) RETURN DISTINCT p.compileunit,d,m

And finally, you have to aggregate d and m as there are more than one for a distinct p.compileunit and so they can appear on a single row. If you don't DISTINCT operator is useless and will be ignore by Neo4j.

When you RETURN your query result, do not forget that there is only 2 dimensions in a RETURN clause table unlike Neo4j where this is an unlimited number of connections and so dimensions.

MATCH (p:ProgNode {compileunit:'Way-Duy_User-Task-Manager', release:2})-[d:RUNS]->(m) RETURN DISTINCT p.compileunit,collect(d),collect(m)

There is a lot of ways to aggregate your data it's up to you but if you want one single row per p.compileunit you have to do it.

3

Thank you. That made sense and worked. Thanks again