使用正则表达式从SPARQL查询中提取信息

Question

I am having a hard time creating a regular expression that extracts the namespaces from this SPARQL query:

SELECT * 
WHERE {
    ?Vehicle rdf:type umbel-sc:CompactCar ;
             skos:subject <http://dbpedia.org/resource/Category:Vehicles_with_CVT_transmission>;
             dbp-prop:assembly ?Place.
    ?Place geo-ont:parentFeature dbpedia:United_States .
}

I need to get:

"rdf", "umbel-sc", "skos", "dbp-prop", "geo-ont", "dbpedia"

I need a expression like this:

\\s+([^\\:]*):[^\\s]+

But the above one does not work, because it also eats spaces before reaching : . What am I doing wrong?

Answer 1

I don't know the details of SPARQL syntax, but I would imagine that it is not a regular language so regular expressions won't be able to do this perfectly. However you can get pretty close if you search for something that looks like a word and is surrounded by space on the left and a colon on the right.

This method might be good enough for a quick solution or if your input format is known and sufficiently restricted. For a more general solution suggest you look for or create a proper parser for the SPARQL language.

With that said, try this:

string s = @"SELECT * 
WHERE {
    ?Vehicle rdf:type umbel-sc:CompactCar ;
    skos:subject <http://dbpedia.org/resource/Category:Vehicles_with_CVT_transmission>;
    dbp-prop:assembly ?Place.
    ?Place geo-ont:parentFeature dbpedia:United_States .
}";

foreach (Match match in Regex.Matches(s, @"\s([\w-]+):"))
{
    Console.WriteLine(match.Groups[1].Value);
}

Result:

rdf
umbel-sc
skos
dbp-prop
geo-ont
dbpedia

Answer 2

So I need a expression like this:

 \\\\s+([^\\\\:]*):[^\\\\s]+ 

But the above one does not work, because it also eats spaces before reaching ":".

The regular expression will eat those spaces, yes, but the group captured by your parenthesis won't contain it. Is that a problem? You can access this group by reading from Groups[1].Value in the Match object returned from Regex.Match .

If you really need the regex to not match these spaces, you can use a so-called look-behind assertion :

(?<=\s)([^:]*):[^\s]+

As an aside, you don't need to double all your backslashes. Use a verbatim string instead, like this:

Regex.Match(input, @"(?<=\s)([^:]*):[^\s]+")