START r=rel(*)
return distinct(type(r))


On Monday, November 25, 2013 2:41:31 AM UTC+5:30, Jacqui Read wrote:
>
> Is there a way to get it to return only distinct relationship types?
>
> On Friday, 30 August 2013 19:35:24 UTC+1, Michael Hunger wrote:
>>
>> then "r" is a collection and you can do:
>>
>> return extract(rel in r : type(rel)) as types
>>
>>
>> Am 30.08.2013 um 19:56 schrieb Karl Grossner <[email protected]>:
>>
>> And how about when the query traverses more than one relationship type, 
>> e.g. [r*0..1]
>>
>> The error there is 'Expected r to be a relationship, but it was a 
>> collection'
>>
>> thanks!
>>
>> On Monday, August 13, 2012 5:45:16 AM UTC-7, Michael Hunger wrote:
>>>
>>> type(r) see 
>>> http://docs.neo4j.org/chunked/milestone/query-function.html#_scalar_functions<http://www.google.com/url?q=http%3A%2F%2Fdocs.neo4j.org%2Fchunked%2Fmilestone%2Fquery-function.html%23_scalar_functions&sa=D&sntz=1&usg=AFQjCNEnN4k7FH7wE6ke72P3dF06q8g3JQ>
>>>
>>> Michael
>>>
>>> Am 13.08.2012 um 14:24 schrieb Murat:
>>>
>>> In a Cypher query I want to return just the type of the relationship, 
>>> and not the whole relationship object? How do I express that? I tried out 
>>> the property syntax as below without success.
>>>  
>>> START n=node(1) 
>>> MATCH n-[r]->b RETURN 
>>> n.name<http://www.google.com/url?q=http%3A%2F%2Fn.name%2F&sa=D&sntz=1&usg=AFQjCNHUdH6df6UT9smHH-LJPYMNtu9epQ>,
>>>  
>>> r.*type*, 
>>> b.name<http://www.google.com/url?q=http%3A%2F%2Fb.name%2F&sa=D&sntz=1&usg=AFQjCNHloW8pFFFUbjT9OluTCIfC4b7SSg>
>>>  
>>> thanks.
>>>
>>>
>>>
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