Get distinct values based on two nodes neo4j - neo4j

Question

There are many 'User' nodes any user can send money to other user.
(:User)-[r:SENT_MONEY]->(:User)
here r has properties
created_at = timestamp()
money_transferred = amount of money transferred
How can I find one users last sent money to other user or last received money from other user in single query.
I have tried this query
MATCH (from:User)-[r:SENT_MONEY]->(to:User)
where (id(from)=1234 OR id(to)=1234)
return max(r.created_at) as sent_at,
r.money_transferred as amount, from.username,to.username
order by last_amount_sent_at DESC
Results are like:
sent_at | amount | from.username | to.username
1408961056 | 20 | user1 | user2
1408961041 | 30 | user2 | user1
1408961028 | 50 | user1 | user3
1408951163 | 20 | user4 | user1
1408951140 | 10 | user1 | user4
By this query user "user1" records with "user2" and "user4" comes twice. It should come single with last transaction between those users like
sent_at | amount | from.username | to.username
1408961056 | 20 | user1 | user2
1408961028 | 50 | user1 | user3
1408951163 | 20 | user4 | user1

Answer 1

Satish,
Here's a query that I think will do what you need.
MATCH (m:User {username : 'user1'})-[r:SENT_MONEY]-(n:User)
WITH m, n, collect(r) AS rs, max(r.created_at) AS p
WITH m, n, filter(x IN rs WHERE x.created_at = p) AS l
RETURN STARTNODE(l[0]), ENDNODE(l[0]), l[0]
For each pair of users you collect the transactions and find the last one, then return the start and end node of that last transaction.
Grace and peace,
Jim
In answer to your further question, you can modify the query like this (for example) to avoid the collection vs element error:
MATCH (m:User {username : 'user1'})-[r:SENT_MONEY]-(n:User)
WITH m, n, collect(r) AS rs, max(r.created_at) AS p
WITH m, n, filter(x IN rs WHERE x.created_at = p) AS o
RETURN m, n, o[0], (o[0]).money_transferred