Yes, I agree.

But for the quantity of data I have until now, the difference is very very 
little.


On Friday, June 6, 2014 6:23:57 PM UTC+2, Michael Hunger wrote:
>
> That's more expensive then you should see it profile
>
> Sent from mobile device
>
> Am 06.06.2014 um 17:32 schrieb Michael Azerhad <[email protected] 
> <javascript:>>:
>
> 456 was well returned. 
> However, I change the query to specify the condition in a where clause : 
> MATCH (user)-[:PARTICIPATES]->(meeting:Meeting {id: "123"})
> WHERE {:User {id: "456"}}-[:PARTICIPATES]->(meeting) 
> RETURN user
>
> Leading to the same result as yours 
> Le 6 juin 2014 à 16:15, Michael Hunger <[email protected] 
> <javascript:>> a écrit :
>
> But then 456 is not returned, or?
>
> Sent from mobile device
>
> Am 06.06.2014 um 14:42 schrieb Michael Azerhad <[email protected] 
> <javascript:>>:
>
> Hi Michael :)
>
> Great, works like a charm.
>
> I reduced it to: 
>
> MATCH (:User {id: "456"})-[:PARTICIPATES]->(meeting:Meeting {id: "123"}), 
> (user)-[:PARTICIPATES]->(meeting)
> RETURN user
>
> Thanks a lot,
>
> Michael
>
> On Friday, June 6, 2014 1:42:29 PM UTC+2, Michael Hunger wrote:
>>
>> MATCH (:User {id: "456"})-[:PARTICIPATES]->(meeting:Meeting {id: "123"})
>> MATCH (user)-[:PARTICIPATES]->(meeting)
>> RETURN user
>>
>> Sent from mobile device
>>
>> Am 06.06.2014 um 13:25 schrieb Michael Azerhad <[email protected]>:
>>
>> MATCH (user)-[:PARTICIPATES]->(meeting:Meeting {id: "123"})
>> RETURN user
>>
>> This query returns all users participating to the Meeting `123`.
>>
>> Now I want to return all users participating to the Meeting `123` *ONLY 
>> if user `456` participates in it*.
>>
>> So concretely, if user1, user2 and user3 participates to the Meeting 
>> `123`, I don't want to return them.
>>
>> However if  user1, *user456* and user3 participates, I want to return 
>> all of them (since `456` exists).
>>
>>
>> How could I achieve this query? 
>> I thought about `WHERE` mixed with `ANY` but didn't succeed.
>>
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