pplication needs to lock the entire database to
> make
>>> the traverse a request for it. But the base operates in real time and the
>>> application should not block the base.
>>> Maybe somehow I can make request to the application by URL? Or some way I
>>> can make
n real time and the
>> application should not block the base.
>> Maybe somehow I can make request to the application by URL? Or some way I
>> can make request to traverse the base without locking?
>>
>> --
>> Vie
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abaseService graphDb = new
> EmbeddedReadOnlyGraphDatabase("/home/sometime/neo4j/data/graph.db");
>
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I found the solution.
Initialization should be done like this:
GraphDatabaseService graphDb = new
EmbeddedReadOnlyGraphDatabase("/home/sometime/neo4j/data/graph.db");
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make request to the application by URL? Or some way I
>> can make request to traverse the base without locking?
>>
>> --
>> View this message in context:
>> http://neo4j-community-discussions.438527.n3.nabble.com/Neo4j-Traverse-Order-By-tp3209366p3242058.html
>
o traverse the base without locking?
>
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> Nabble.com.
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Just my two cents: I actually wrote sort of a similar plugin: it implemented
a slightly modified 'breadth-first' logic, where nodes were added to
PriorityQueue (rather than simple queue, as in original ordering class).
Hope this'll help... )
On 29 July 2011 17:02, Jim Webber wrote:
> Hey,
>
> Re
Hey,
Retrieving "popular" nodes not directly possible with the REST API.
Even in Java, you'd have to sort your result set manually since you don't know
ahead of time how the search is going to traverse your graph.
It is only once you have your result set (which may be large) that you can
sort
Hi!
For example, we have social graph: http://www.donskov.net/graph.jpg
Nodes is humans. If the nodes have a relation - so they are friends.
I need to get all friends friends and all friends friends friends etc. on
node 4 (see example).
I post this:
{
"order": "breadth_first",
"u
and 5) - is are friends of our start node (4), so node 10
should be the first node than 5 (because nodes 1 and 9 have one relation
from node(4)'s friends).
How do I change the query to get this sorted?
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