At the end I decide to take a shortcut that also might help explain what
I want:
|private URI rdfLiteralToUri(String literal) { int xsdIndex =
literal.lastIndexOf("^^"); if (xsdIndex == -1) { throw new
IllegalArgumentException("Not valid literal format"); } String uriStr =
literal.substring(0, xsdIndex); String type = literal.substring(xsdIndex
+ 2, literal.length()); if
(!XSDDatatype.XSDanyURI.getURI().equals(type)) { throw new
IllegalArgumentException("Not valid literal type"); } return
URI.create(uriStr); } |
On 11/01/2017 16:42, George News wrote:
> On 11/01/2017 16:25, Chris Dollin wrote:
>> On 11/01/17 14:55, George News wrote:
>>> Hi,
>>>
>>> I have this literal:
>>> http://hola^^http://www.w3.org/2001/XMSchema#anyURI
>> What do you mean by "have"? A String value, a Literal
>> value, or what?
> I want to get http://hola as a String or as a URI. Better as a URI, but
> only if it is conforming with the type linked.
>
>
>>> And I want to create a URI from it. Is there any way to do so?
>> And do you want an actual URI object or just its spelling?
>>
>> Because you can get the spelling of the URI using getLexicalForm.
> Literal a = (Literal)
> ResourceFactory.createTypedLiteral("http://hola^^http://www.w3.org/2001/XMSchema#anyURI";,
> XSDDatatype.XSDanyURI);
> System.out.println(a.getDatatype());
> System.out.println(a.getLexicalForm());
> System.out.println(a.getDatatypeURI());
> System.out.println(a.getString());
>
> Output:
> Datatype[http://www.w3.org/2001/XMLSchema#anyURI -> class java.net.URI]
> http://hola^^http://www.w3.org/2001/XMSchema#anyURI
> http://www.w3.org/2001/XMLSchema#anyURI
> http://hola^^http://www.w3.org/2001/XMSchema#anyURI
>
>
> None of them returns "http://hola"; which is the actual URI.
>
>> Chris
>>
