mmm thank you Michael,
i will try this out.

i was also getting 0 rows because i noticed that i previously created
relationshpis between nodes by limiting the results, so it was not said
that, between two nodes, there would have been the two types of
relationships I was looking for.


On Sun, Aug 10, 2014 at 4:05 PM, Michael Hunger <
[email protected]> wrote:

> That's probably because you'd want to use two different _m's_ otherwise
> the first m is reused for the second match.
>
> MATCH (n:TYPE)
> WHERE n.id = 'hellyeah'
> OPTIONAL MATCH (n)-[r:P]-(m1)
> OPTIONAL MATCH (n)-[s:P2]-(m2)
> RETURN n,r,s,m1,m2
>
> You can also optimize this a bit to do the second bit of matches not for
> each of the first
> Something like this:
>
> MATCH (n:TYPE)
> WHERE n.id = 'hellyeah'
> OPTIONAL MATCH (n)-[r:P]-(m1)
> WITH n, collect({rel:r, node:m1}) as neighbours
> OPTIONAL MATCH (n)-[s:P2]-(m2)
> WITH collect({rel:s, node:m2}) + neighbours as neighbours
> UNWIND neighbours  as nb
> RETURN nb
>
>
>
>
>
> On Fri, Aug 8, 2014 at 1:04 AM, gg4u <[email protected]> wrote:
>
>> Hi folks!
>>
>> so, today I was experimenting on obtaining a linear combination out of
>> multiple types of properties.
>> Suppose you have multiple REL types, each with weighted properties.
>>
>> I am trying to obtain a collection? of first neighbors to a node, ordered
>> by the sum of two weights.
>>
>> Here's what i've done:
>>
>> MATCH (n:TYPE)
>> WHERE n.id = 'hellyeah'
>> OPTIONAL MATCH n-[r:P]-m
>> OPTIONAL MATCH n-[s:P2]-m
>> RETURN n,r,s,m
>>
>> with this query, i get a table BUT all values of s are set as
>> NULL
>> could you help me understand
>> why?
>>
>> thank you!
>>
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