SPARQLWrapper 1.5.2
SPARQLWrapper是一個SPARQL端點接口到Python。它是圍繞一個SPARQL服務的包裝。它有助於在創建查詢URI和，可能的話，轉換的結果成更易於管理的格式。該SPARQLWrapper包下W3C許可證授權。在這裡，您對如何使用圖書館在你的Python程序的例子：從SPARQLWrapper進口SPARQLWrapper，JSONSPARQL = SPARQLWrapper（“http://dbpedia.org/sparql”）sparql.setQuery（“”“＆NBSP;...
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