我正在尝试列出所有完成3件以上工作的所有作者-DBpedia Sparql [英] I am trying to get list of all the authors who have had more than 3 piece of work - DBpedia Sparql

问题描述

我正在尝试列出所有完成3件或以上工作的作者(在DBpedia中).

I am trying to get list of all the authors who have had 3 or more piece of work done (in DBpedia).

我的示例可以在以下计算机上运行: http://dbpedia.org/sparql

my example can be run on : http://dbpedia.org/sparql

select (count(?work) as ?totalWork), ?author
Where
{
  ?work dbo:author ?author.
}
GROUP BY ?author

我让每个作者完成全部工作.但是，当我尝试过滤以仅显示具有3件以上作品的作者列表时.我收到错误消息:

I get each authors total amount of piece of work done. But when I try to filter to show only list of author that have more than 3 piece of work. I get error:

我尝试使用HAVING关键字或使用FILTER关键字.

I tried HAVING keyword or using FILTER keyword.

使用过滤器

select (count(?work) as ?tw), ?author
Where
{
  ?work dbo:author ?author.
  FILTER (?work > 3).
}
GROUP BY ?author

error: Virtuoso 22023 Error VECDT: SR066: Unsupported case in CONVERT (INTEGER -> IRI_ID)

使用HAVING关键字

select (count(?work) as ?tw), ?author
Where
{
  ?work dbo:author ?author.
}
GROUP BY ?author
HAVING (?tw > 3)

Virtuoso 37000 Error SP031: SPARQL compiler: Variable ?tw is used in the result set outside aggregate and not mentioned in GROUP BY clause

推荐答案

使用HAVING是正确的，但是存在

Using HAVING is correct, but there is a limitation in SPARQL with indirectly referring to aggregates.

此作品有效:

SELECT (count(?work) as ?tw) ?author
WHERE
{
  ?work dbo:author ?author.
}
GROUP BY ?author
HAVING (count(?work) > 3)

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