possible bug

[Javad Karabi]

http://screencloud.net/v/7ddO

[Javad Karabi]

anyone come across anything like this?

On Thursday, February 6, 2014 4:02:59 PM UTC-6, Javad Karabi wrote:

http://screencloud.net/v/7ddO

[Mark Needham]

Can you paste the whole query? I know we had some problems in 2.0.0 with that, which version are you using?

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[Javad Karabi]

mark, sure. the query is below.

fwiw, the problem is caused when "other.birth_year as other_birth_year," is included.

when "other.birth_year as other_birth_year," is removed, it executes fine.

MATCH (me:Member {id: 123})-[:preferred_store]->(s)<-[:preferred_store]-(other)-[:ordered]->()<-[:product]-(sv:StyleVariant)

WITH other.birth_year as other_birth_year, sv.id as style_variant,

    count(s) +

      CASE HAS(me.birth_year) AND HAS(other.birth_year)

        WHEN true THEN

          10 - ABS(me.birth_year - other.birth_year)

        ELSE

          0

        END AS score

WITH SUM(score) as score_for_style_variant, style_variant ORDER BY score_for_style_variant DESC LIMIT 50

RETURN collect(style_variant)

[Mark Needham]

Does this version do the job?

MATCH (me:Member {id: 123})-[:preferred_store]->(s)<-[:preferred_store]-(other)-[:ordered]->()<-[:product]-(sv:StyleVariant)

WITH other.birth_year as other_birth_year, other, sv.id as style_variant, s, me

WITH other_birth_year, style_variant,

    count(s) +

      CASE HAS(me.birth_year) AND HAS(other.birth_year)

        WHEN true THEN

          10 - ABS(me.birth_year - other.birth_year)

        ELSE

          0

        END AS score

WITH SUM(score) as score_for_style_variant, style_variant ORDER BY score_for_style_variant DESC LIMIT 50

RETURN collect(style_variant)

[Javad Karabi]

hmm, that does seem to work. 

what do you suppose the issue is?

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[Mark Needham]

Good question, I'm not sure. Here's some attempts to figure it out:

MATCH (me:Member)

RETURN me.id AS id, COUNT(*) + CASE me.id > me.name WHEN true then 0 ELSE 5 END

Unknown identifier `  UNNAMED彟븥ۚ覴ण`.
Unknown identifier `  UNNAMED̫묖ᩊ⚞䋕`.
Unknown identifier `me`.

Without count:

MATCH (me:Member)

RETURN me.id AS id, CASE me.id > me.name WHEN true then 0 ELSE 5 END

Returned 0 rows in 120 ms

Comparing me.id to me.id:

MATCH (me:Member)

RETURN me.id AS id, COUNT(*) + CASE me.id > me.id WHEN true then 0 ELSE 5 END

Returned 0 rows in 122 ms

COUNT(me) instead of COUNT(*):

MATCH (me:Member)

RETURN me.id AS id, COUNT(me) + CASE me.id > me.name WHEN true then 0 ELSE 5 END

Unknown identifier `  UNNAMED⤂вぐ짃Ǹ`.
Unknown identifier `  UNNAMED᳽퓒讈埂↌`.
Unknown identifier `me`.

Michael, Andres, Wes, Stefan any ideas? 

[Mark Needham]

If I explicitly specify me.name then it's ok as well:

MATCH (me:Member)

RETURN me.id AS id, me.name, COUNT(*) + CASE me.id > me.name WHEN true then 0 ELSE 5 END

Returned 0 rows in 91 ms

It's something around adding to aggregates - you don't seem to be able to reference any identifiers unless you already have them earlier in your RETURN clause.

[Javad Karabi]

any idea why the (chinese?) characters are being displayed? i guess its a sort of 'anonymous' relationship, so neo4j decides to name it using chinese symbols lol

[Mark Needham]

Yeh any unnamed elements (nodes/rels) in a query get given a name like that - it uses a random string generator to come up with the name. 

[Michael Hunger]

Random.nextString() hopefully resolved soon.

Michael