rdf - Virtuoso 中的 STRAFTER SPARQL 1.1 函数

Question

是否STRAFTER从 SPARQL 1.1 工作在 Virtuoso(2012 年 6 月 6.04 版)？

尝试

PREFIX qb:  <http://purl.org/linked-data/cube#>
PREFIX iodp: <http://data.oceandrilling.org/core/1/>
PREFIX sdmx-dimension:  <http://purl.org/linked-data/sdmx/2009/dimension#>
SELECT ?p (strafter(str(?p), "http://data.oceandrilling.org/janus/") AS ?pshort)
FROM <http://data.oceandrilling.org/janus/>
WHERE {
   <http://data.oceandrilling.org/janus/ngr_section201_1226> qb:observation ?observation .
   ?observation ?p ?o .
   FILTER (regex(?p, "janus")) .
   FILTER (!regex(?p, "leg")) .
   FILTER (!regex(?p, "site")) .
   FILTER (datatype(?o) = <http://www.w3.org/2001/XMLSchema#decimal>) .
}

在 http://data.oceandrilling.org/sparql引发 SPARQL 编译错误。

Answer 1

不，它不支持 sraft，您需要构建 6.1.5 see the release notes .