Hello,
thanks for encouraging me, just got it working as you said,
Jaana
Andy Seaborne kirjoitti 6.5.2023 18:55:
On 05/05/2023 11:50, [email protected] wrote:
Thanks for your answer, but I still don't undestand how to combine
those two queries.
I'm not saying they can simply be joined together.
If the FROM tilasto:$STAT is avoided, and the other FROM can be
removed and GRAPH used then there is probably a single query. That
depends on the data and data model.
Andy
If I put them like this jena-fuseki-UI doesn't accept line "{
graph tilasto:?ng", because ?ng comes from the 1st subquery.
It may take a LATERAL join but
BIND ( make IRI with tilasto: and ?ng AS ?gn )
GRAPH ?gn
is a possible way to modify the query.
It does all depend on the data and data model which you are the expert
for.
SELECT *
WHERE {
{
{
GRAPH stat:outputchannel
{
?subject outchl:refersToTable ?t_id .
?subject outchl:refersToNamedGraph ?ng .
?subject outchl:hasOutputFileId ?o_id.
filter (regex(?o_id,"of_001"))
}
}
}
{ graph tilasto:?ng
{
?pxfile pxt:tableId "?t_id".
?pxfile pxt:isPresentationOf ?cube.
?cube dc:description ?title_fi.
?cube cubemeta:refersToOutputFile ?of.
}
}
{ graph stat:outputchannel
{
?subject outchl:hasOutputFileId ?of.
?subject outchl:refersToOutputChannel ?channel_prefix .
?subject outchl:refersToTable ?t_id .
?subject outchl:refersToNamedGraph ?ng .
?subject outchl:hasOutputFileId ?o_id.
}
}
{ graph stat:tabAdmin
{
?kanta outchl:hasOutputChannelId ?channel_prefix .
?kanta outchl:directoryPathRoot ?directoryPathRoot .
}
}
}
Can you help ?
Br, Jaana
Andy Seaborne kirjoitti 4.5.2023 13:14:
On 04/05/2023 07:18, [email protected] wrote:
Andy Seaborne kirjoitti 4.5.2023 00:24:
Do you have to use FROM in the second query?
I don't know how to present it because in the 2nd query I'm querying
three named graphs, where the third one ($STAT) should be replaced
with the result of the 1st query (?ng)
But do you need a merged graph or can you use GRAPH? FROM of multiple
graphs may be a significant cost. (It precludes TDB executing more
directly on basic graph patterns.)
With GRAPH ?g you can apply a condition to the ?g.
And that means you can combine the queries which might help - to
know,
needs an experiment.
---
3000 queries in 5 mins is 100ms a query, including client and server
overheads.
Are you doing the 3000 queries in parallel? A bit of parallelism
might
save elapsed time (start with parallel = 2).
Andy
Br, Jaana
On 03/05/2023 17:58, [email protected] wrote:
Hello,
I have the two queries below which I run from my code so that the
1st query returns about 3000 ?ng and ?t_id pairs which will then
be used in the second query in the place of $STAT and $RDF_ID. So
I'm calling the second query in a loop about 3000 times.
I've noticed that it is time consuming; it takes about 4-5
minutes. How could I combine the queries so that I could get all
the information by just one call ?
$PREFIXS
SELECT *
WHERE
{
{
GRAPH stat:outputchannel
{
?subject outchl:refersToTable ?t_id .
?subject outchl:refersToNamedGraph ?ng .
?subject outchl:hasOutputFileId ?o_id.
filter (regex(?o_id,"of_001"))
}
}
}
and
$PREFIXS
SELECT *
FROM stat:outputchannel
FROM stat:tabAdmin
FROM tilasto:$STAT
WHERE {
?pxfile pxt:tableId "$RDF_ID".
?pxfile pxt:hasStatus ?status.
?pxfile pxt:hasFrequency ?frequency.
?pxfile pxt:isPresentationOf ?cube.
?cube dc:description ?title_fi.
?cube dc:language ?language.
?cube cubemeta:hasStatisticalProgramme
?statisticalProgram.
?cube cubemeta:lastUpdated ?lastUpdated.
?cube cubemeta:refersToOutputFile ?of.
?subject outchl:hasOutputFileId ?of.
?subject outchl:refersToOutputChannel ?channel_prefix .
?kanta outchl:hasOutputChannelId ?channel_prefix .
?kanta outchl:directoryPathRoot ?directoryPathRoot .
?kanta a outchl_ont:OutputChannel .
filter(?language ="fi"^^xsd:language)
filter (lang(?title_fi) = "fi")
filter ( langMatches(lang(?directoryPathRoot),"fi") )
}
Br jaana M
