[SQL/Match Question] Get node count between + edge data in a single query - "optimize" ?

[matanshukry]

OrientDB Version: 3.0.20

Java Version: 1.8.0_191

OS: Windows 10

Assuming I have this model:

Person (userId String)
Friend (from person to person) (status: 1=requested,2=approved)

I need to create a query that will find all Person that will match 'John' (Lucene full text), and will return (a) the friend data between them and person 'Bar', and (b) The amount of friends that person and 'Bar' has in common.

The current query I have is:

SELECT (MATCH {class:Person, where:(SEARCH_CLASS("John") = true)}.bothE("Friend"){as:friend, where:(status=2)}.bothV("Person"){where:(userId="bulu")} RETURN friend:{*}) as friend, (MATCH {class:Person, where:(SEARCH_CLASS("John") = true)}.bothE("Friend"){where:(status=2),as:fs1}.bothV("Person"){as:friend, where:($matched != $currentMatch)}.bothE("Friend"){where:(status=2 AND $matched.fs1 != $currentMatch)}.bothV("Person"){where:(userId="bar" AND $matched.friend != $currentMatch)} RETURN COUNT(friend) as cnt) as count

Which technically is working, but:
(1) It looks like a really bad query, performance-wise. I have to repeat few things in each sub-query (such as SEARCH_CLASS). Anyway to avoid that?
(2) The result contains 2 columns, 'friend' and 'count', but in each of them I have an array, and in the count I even have a nested json with 'cnt'. Is there a way to avoid all that, so that in the first column/cell I will simply have the json data of 'friend', and in the second I will simply have an integer - the count?

[luigidellaquila]

Hi @matanshukry

You can try the following:

SELECT friend:{*}, nFriends FROM (
  MATCH 
   {class:Person, where:(SEARCH_CLASS("John") = true), as:p1}
   .bothE("Friend"){as:friend, where:(status=2)}
   .bothV("Person"){where:(userId="bulu")} 
   .bothE(){as:otherFriend}.bothV(Friend){where:($matched.p1 = $currentMatch)}
  RETURN friend, count(otherFriend) as nFriends
  GROUP BY friend
)

I didn't test it, so there is a chance that it has typos, but I hope you got the idea

Thanks

Luigi

[luigidellaquila]

sorry, here's the right query

SELECT friend:{*}, nFriends FROM (
  MATCH 
   {class:Person, where:(SEARCH_CLASS("John") = true), as:p1}
   .bothE("Friend"){as:friend, where:(status=2)}
   .bothV("Person"){where:(userId="bulu")} 
   .both("Friend"){as:otherFriend}.both("Friend"){where:($matched.p1 = $currentMatch)}
  RETURN friend, count(otherFriend) as nFriends
  GROUP BY friend
)

Thanks

Luigi

[matanshukry]

@luigidellaquila Awesome! I really liked the way you did it, great!

One question though: why do we need the SELECT and the GROUP BY? the match it self seems to return the same data (with very little edit of a field in the RETURN)

MATCH 
	{class:Person, where:(SEARCH_CLASS("John") = true), as:p1}
	.bothE("Friend"){as:friend, where:(status=2)}
	.bothV("Person"){where:(userId="bulu")} 
	.both("Friend"){as:otherFriend}.both("Friend"){where:($matched.p1 = $currentMatch)}
RETURN friend:{*}, count(otherFriend) as nFriends

[luigidellaquila]

Yes, it should work (just add a GROUP BY).
My only doubt is if the grouping on the expanded "friend" will execute correctly (I have to double-check it)

Thanks

LUigi

[matanshukry]

@luigidellaquila - that's the thing I don't understand - why do we need to add a GROUP BY? the match I wrote is meant to be used alone, without a select at all. There should only be one friend anyway, due to how the match was written.. no?

Also, I noticed after we find the target person, {where:(userId="bulu")}, we keep looking for Friend->person (as otherFriend), and again (to match current match). Shouldn't both these connections have status=2 on them too though? hence each .both("Friend") should actually be .bothE("Friend"){where:(status=2)}.bothV("User") ?

[luigidellaquila]

The GROUP BY is needed just because you have an aggregation function (count()).

About the status=2, yes, if that's a requirement, then your solution is correct

Thanks

Luigi

[matanshukry]

@luigidellaquila How come it works without the GROUP BY then? I ran the Match only and it returned the correct result, without the select/group by..

Regarding status=2: the value 2 is coming as a parameter actually. Is there anyway to avoid using it 3 times? (in each where)

[luigidellaquila]

Hi @matanshukry

The execution planner is smart enough to add the GROUP BY automatically in some cases, but I always suggest to set it explicitly to make the query more clear.

You can pass named parameters to the query as :paramName (eg. via Java API) as follows:

String query = "";
query += "MATCH ";
query += "	{class:Person, where:(SEARCH_CLASS("John") = true), as:p1}";
query += "	.bothE("Friend"){as:friend, where:(status = :status)}";
query += "	.bothV("Person"){where:(userId="bulu")} ";
query += "	.bothE("Friend"){where:(status = :status)}.bothV(){as:otherFriend}";
query += "	.bothE("Friend"){where:(status = :status)}.bothV(){where:($matched.p1 = $currentMatch)}";
query += "RETURN friend:{*}, count(otherFriend) as nFriends";

Map<String, Object> params = new HashMap<>();
params.put("status", 2);

OResultSet resultSet = db.query(query, params);

Thanks

Luigi

[matanshukry]

@luigidellaquila Awesome, that solves all my problems and questions and then some! Thanks :)

[matanshukry]

@luigidellaquila Spoke a bit too soon;
When there are no mutual friends (nFriends == 0), I'm not getting any results. Even in such a case, I still need to get the 'friend'. How do I do that?

[luigidellaquila]

Hi @matanshukry

You can add an optional:true to the items you expect do not exist:

query += "	.bothE("Friend"){where:(status = :status), optional:true}.bothV(){as:otherFriend, optional:true}";
query += "	.bothE("Friend"){where:(status = :status), optional:true}.bothV(){where:($matched.p1 = $currentMatch)}";
q

Thanks

Luigi

[luigidellaquila]

well, actually you can write it as follows, that is much better:

MATCH 
	{class:Person, where:(SEARCH_CLASS("John") = true), as:p1}
	.bothE("Friend"){as:friend, where:(status = :status)}
	.bothV("Person"){where:(userId="bulu")} 
	.bothE("Friend"){where:(status = :status), optional:true}.bothV(){as:otherFriend, optional:true}
	.bothE("Friend"){where:(status = :status), optional:true}.bothV(){as:p1}
RETURN friend:{*}, count(otherFriend) as nFriends

(see the {as:p1} instead of {where:($matched.p1 = $currentMatch)})

Thanks

Luigi