Hello,
I have a question about querying 'hierarchical' data in a graph.
Let's assume I have a Person( name ) class, an EdgeClass 'owns', and a
Bicycle( color, type ) class.
Let's assume than John owns 2 bicycles: ( 'red', 'mountain bike' ) and (
'blue', 'recumbent bicycle' )
I found out about the expand function, so in order to get a person with the
2 types of bicycles, I could query:
select name,
outE()[ @class = 'owns' ].inV().type as bicycletypes
from Person
And if I do
select name,
outE()[ @class = 'owns' ].inV().type as bicycletypes
from Person
I get something like:
{
"result": [
{
"@type": "d",
"@rid": "#-2:1",
"@version": 0,
"name": "John",
"bicycletypes": [ 2 ]
}
],
"notification": "Query executed in 0.03 sec. Returned 1 record(s)"
}
But I would like to have something like:
{
"result": [
{
"@type": "d",
"@rid": "#-2:1",
"@version": 0,
"name": "John",
* "bicycletypes": [ "recumbent bicycle", "mountain bike" ]*
}
],
"notification": "Query executed in 0.03 sec. Returned 1 record(s)"
}
And even better would be:
{
"result": [
{
"@type": "d",
"@rid": "#-2:1",
"@version": 0,
"name": "John",
* "bicycles": [ { "color": "red", "type": "recumbent bicycle"
}, *
*{ "color": "red", "type": **"mountain bike" } *
* ]*
}
],
"notification": "Query executed in 0.03 sec. Returned 1 record(s)"
}
When I try to get a similar result, for example with the following query:
select *{* "name": name, "bicycles": outE()[ @class = 'owns' ].inV() *}* as
expandedperson
from Person
I get
{"name":"_NOT_PARSED_","bicycles":"_NOT_PARSED_"}
or something else I tried
select *, expand( outE()[ @class = 'owns' ].inV() ) as bicycles
from Person
doesn't work either.
*So, my question isn (if possible using OrientDB SQL queries)*
- *can something like this be done in a single query, and if so, how? *(I
guess not currently?)
-
*Are there plans to make such a thing possible in the future? I would
really love to be able to do these things. Something similar to the
highlighted syntax would rock :) *
-
*Are there 'workarounds' that I can use for now (with the least possible
amount of queries)? *
Thanks again for all your help getting me started.
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