public TextResource text(XPathQuery path) throws Exception {
		String uri = path.eval(this, String.class);
		return text(uri);
	}
public TextResource text ( XPathQuery path ) throws Exception { String uri = path . eval ( this , String . class ) ; return text ( uri ) ; }
Execute the given path query on the XML, GET the returned URI expecting text/* as content

@param path path to the URI to follow, must be a String QName result
@return a new resource, as a result of getting it from the server in text format
@throws Exception