sparql-提取URI的最后一部分
我有来自不同域的URI列。例如，
http://comicmeta.org/cbo/category
http://purl.org/dc/terms/hasVersion
http://schema.org/contributor
依此类推。我想在每个此类URI上的最后一个斜杠'/'之后提取最后一部分，即，字符串。
上面的URIS列表中的预期结果：
category
hasVersion
contributor
如何编写 generic sparql查询以从任何给定的URI中提取最后一部分？
这就是我到目前为止尝试的：
SELECT distinct ?s ?x WHERE {
?s ?p ?o .
BIND (STRBEFORE(STRAFTER(STR(?s),"/"), " ") as ?x) .
#To extract the part after the slash '/' and before the end of string indicated by a space ' '.
}
但是，这只返回空字符串“”。
我该如何完成这项工作？有人可以帮我吗？
I have a column of URIs from different domains. Example,
http://comicmeta.org/cbo/category
http://purl.org/dc/terms/hasVersion
http://schema.org/contributor
and so on. I want to extract the last part, i.e, the string after the last slash '/' on each such URI.
Expected results on the above list of URIs:
category
hasVersion
contributor
How do I write a generic SPARQL query to extract this last part from any given URI?
This is what I have tried so far:
SELECT distinct ?s ?x WHERE {
?s ?p ?o .
BIND (STRBEFORE(STRAFTER(STR(?s),"/"), " ") as ?x) .
#To extract the part after the slash '/' and before the end of string indicated by a space ' '.
}
But, this only returns empty strings "".
How can I make this work? Can someone help me with this?
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使用
替换是这样：这仅在最后一个
/字符之后找到的零件代替整个字符串。但是请注意，由于您的示例，并非所有词汇量都使用/作为定界符；有些还使用＃。http://www.w3.org/1999/02/02/22-rdf-syntax-ns#type将变成22-rdf-syntax-ns＃代码>如果您不想要它，则可以使用一些更复杂的东西：
这通常是根据通常是有效的XML名称来选择最长的部分。
Using
REPLACEis the way:This replaces the whole string with only the part found after the last
/character. Note however that, due to your examples, not all vocabularies use/as the delimiter; some also use#. Something likehttp://www.w3.org/1999/02/22-rdf-syntax-ns#typewill be turned into22-rdf-syntax-ns#typeIf you do not want that, you could use something a bit more complicated:
This selects the longest part from the end based on what usually is a valid XML name.