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How to use SPARQL query to get field and field count at the same time?


How to use SPARQL query to get field and field count at the same time?

By : vikas chandra
Date : November 23 2020, 09:01 AM
help you fix your problem I am new to Semantic Web and SPARQL. I am making a RDFS of concert. It has songs entries like: ,
code :
SELECT ?songs (COUNT(?artist) AS ?count)  
WHERE
{ 
  ?s rdfs:label ?songs .
  ?s rdfs:Class con:Repertoire.
  ?s con:performedBy ?artist.
  ?artist rdfs:Class con:Artist .
  filter( ?count > 1 )
}
SELECT ?songs (COUNT(?artist) AS ?count)  
WHERE
{ 
  ?s rdfs:label ?songs .
  ?s rdfs:Class con:Repertoire.
  ?s con:performedBy ?artist.
  ?artist rdfs:Class con:Artist .
}
GROUP BY ?songs
HAVING (?count > 1 )


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How to count a field and this same field but filtered, in one query in Django?

How to count a field and this same field but filtered, in one query in Django?


By : user3359301
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , I'm not sure how to do it in one query, but consider something like this:
code :
global_votecount = Vote.objects.all().values('user', 'user__username').annotate(votesTotal=Count('user__username')).order_by('-votesTotal')

import datetime
today_min = datetime.datetime.combine(datetime.date.today(), datetime.time.min)
today_max = datetime.datetime.combine(datetime.date.today(), datetime.time.max)

votesToday = dict(Vote.objects.filter(date__range=(today_min, today_max)).values_list('user__username').annotate(votesToday=Count('user__username')))


for item in global_votecount:
    item['votesToday'] = votesToday[item['user__username']]
Save Followers count in a field or query each time if needed

Save Followers count in a field or query each time if needed


By : user3671176
Date : March 29 2020, 07:55 AM
will help you Well, if you want to know what is right, there is only one answer.
MySQL Query - Update field with order/count but start at 1 again based on another field

MySQL Query - Update field with order/count but start at 1 again based on another field


By : Beichen Yang
Date : March 29 2020, 07:55 AM
help you fix your problem I assume you have an error in your desired result. Sortorder for Product 9 and 10 should be 4 and 5, right?
Here's how you can do it:
code :
UPDATE t
JOIN (

    SELECT
    t.*
    , @rc := IF(@prevpc != product_category OR @prevv != vendor_id, 1, @rc + 1) AS so
    , @prevpc := product_category
    , @prevv := vendor_id
    FROM
    t
    , (SELECT @prevpc := NULL, @prevv := NULL, @rc := 0) var_init_subquery
    ORDER BY product_category, vendor_id, product_name

) sq ON t.product_name = sq.product_name
SET t.sortorder = sq.so;
DBPedia: SPARQL query on field starting with literal

DBPedia: SPARQL query on field starting with literal


By : user2662139
Date : March 29 2020, 07:55 AM
I wish this helpful for you For DBpedia, the best option might be to use the bif:contains full-text search pseudo property:
code :
SELECT ?company {
  ?company a dbo:Company.
  ?company foaf:name ?name.
  ?name bif:contains "Novartis"@en.
}
SELECT ?company {
  ?company a dbo:Company.
  ?company foaf:name ?name.
  FILTER strStarts(?name, "Novartis")
}
SELECT ?company {
  ?company a dbo:Company.
  ?company foaf:name ?name.
  FILTER contains(lCase(?name), lCase("Novartis"))
}
SPARQL query for skos:related field?

SPARQL query for skos:related field?


By : Jouni Mustamo
Date : March 29 2020, 07:55 AM
wish help you to fix your issue As Jeen suggests, finding out more about SPARQL is a good idea. As an initial reference, this is pretty easy:
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