Yes this works.

Thanks


On Thursday, January 26, 2017 at 9:23:22 AM UTC+1, Luigi Dell'Aquila wrote:
>
> Hi Alberto,
>
> please try to replace this
>
> OSequenceLibrary sequenceLibrary = access.graph
> .getRawGraph().getMetadata().getSequenceLibrary();
>
> with this
>
> OMetadata md = access.graph.getRawGraph().getMetadata();
> md.reload();
> OSequenceLibrary sequenceLibrary = md.getSequenceLibrary();
>
> Maybe it's just a matter of caching
>
> Thanks
>
> Luigi
>
>
> 2017-01-26 0:07 GMT+01:00 Maverick <[email protected] <javascript:>>:
>
>> Each time this code is executed, a new sequence is created with the same 
>> name, and so the method always returns 1 as next number. Why?
>>
>>
>> public long next( String seqName ) {
>>
>> OSequenceLibrary sequenceLibrary = access.graph
>> .getRawGraph().getMetadata().getSequenceLibrary();
>>
>> OSequence seq = sequenceLibrary.getSequence( seqName );
>>
>> if( seq == null ) {
>>
>> seq = sequenceLibrary.createSequence(seqName, SEQUENCE_TYPE.ORDERED, new 
>> OSequence.CreateParams().setStart(0L));
>>
>> }
>>
>> return seq.next();
>>
>> }
>>
>>
>> I'm using graphdb v.2.2.13
>>
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>
>

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