xsd - How can you remove the XML schema datattype from sparql query? -

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मैं एक फ़ाइल पर एक स्पार्किल क्वेरी चल रहा है जिसमें 

  & rdquo; user rdf: about = "# आरजे" & gt; & Lt; hasName rdf: datatype = "http://www.w3.org/2001/XMLSchema#string" & gt; आरजे & lt; / hasName & gt; & Lt; / उपयोगकर्ता & gt;

मुझे केवल नाम, 'आरजे' वापस करना है, लेकिन जब मैं अपनी क्वेरी दर्ज करता हूं 

  SELECT? Name FROM & lt; example.com & gt; जहां (असाइन करें: आरजे असाइन करें: हैनाम नाम? }

जहां सही नाम स्थान है, मैं उसे वापस लौटाता हूं: 

  "आरजे" ^^ & lt; http: //www.w3.org/2001 / XMLSchema # स्ट्रिंग & gt; 
 
 
 
 
 
 
 
 
 
 
 
 
 आप ऐसा कर सकते हैं यह SPARQL कार्यान्वयन पर निर्भर करता है जो आप उपयोग कर रहे हैं। SPARQL 1.0 के तहत यह संभव नहीं है, हालांकि अब जिसकी व्यापक रूप से मार्च 2013 में डब्लू 3 सी सिफारिश बनने वाले अधिकांश कार्यान्वयनों का समर्थन किया जा सकता है: 
 
  SELECT (STR (? Name) AS? StringName) से & lt; example.com & gt; मूल रूप से एक परियोजना अभिव्यक्ति आपको किसी भी मान्य SPARQL अभिव्यक्ति का उपयोग करने की अनुमति देती है, जो कि आप वेरिएबल्स के आधार पर एक नया मान की गणना करने के लिए कहीं और का उपयोग कर सकते हैं। पहले बाध्य।

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