Hi Michael,
I am confused with your answer. I am trying to union the first 3 MATCH
queries and throw the result "Users" into
MATCH p = allShortestPaths((n)-[*]-(Users)) RETURN COUNT(*)
to find the shortest paths.
On Saturday, June 21, 2014 6:22:37 PM UTC-4, Michael Hunger wrote:
>
> The final query is:
>
> MATCH (n:User{id:'2'}) - [:WENT_TO] -> (bar) <-[: WENT_TO] -(m3) WHERE
> n.age=m3.age RETURN m3 AS Users
> MATCH p = allShortestPaths((n)-[*]-(Users)) RETURN COUNT(*)
>
>
> not
>
> MATCH (n:User{id:'2'}) - [:WENT_TO] -> (bar) <-[: WENT_TO] -(m3) WHERE
> n.age=m3.age RETURN m3 AS Users
>
>
> Michael
>
> Am 21.06.2014 um 22:13 schrieb Alx <[email protected] <javascript:>>:
>
> Sure! Here are the 3 union queries with the red one being the final query.
> I know I can put the final query inside each individual query but I guess
> this is not very efficient. Thanks for the help.
>
> MATCH (n:User{id:'2'}) - [:LOCATED_IN|:HAS_FAX_NUMBER|:HAS_PHONE_NUMBER]
> -> () <-[:LOCATED_IN|:HAS_FAX_NUMBER|:HAS_PHONE_NUMBER]-(m1) WHERE
> m1.type='Professional' RETURN m1 AS Users
> UNION
> MATCH (n:User{id:'2'}) - [:TEXTED] ->() < - [: TEXTED] -(m2) MATCH p =
> allShortestPaths((n)-[*]-(m2)) RETURN m2 AS Users
> UNION
> MATCH (n:User{id:'2'}) - [:WENT_TO] -> (bar) <-[: WENT_TO] -(m3) WHERE
> n.age=m3.age RETURN m3 AS Users
> MATCH p = allShortestPaths((n)-[*]-(Users)) RETURN COUNT(*)
>
>
> On Saturday, June 21, 2014 11:05:31 AM UTC-4, Michael Hunger wrote:
>>
>> Perhaps you can share your query with us, perhaps it is easier to rewrite
>> it to something else?
>>
>> In general right now it is not possible to do anything else with the
>> results of a union query.
>> Michael
>>
>> Am 19.06.2014 um 21:58 schrieb Alx <[email protected]>:
>>
>> I want to use UNION to remove the duplicates from 3 MATCH queries and
>> apply a query on the results. Is it possible? If not what 's the workaround?
>>
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