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Error in STRAFTER() example?

From: Steve Harris <steve.harris@garlik.com>
Date: Mon, 5 Nov 2012 12:34:08 +0000
Message-Id: <4AD20498-FF98-4FC3-9E10-F17513C6AEE9@garlik.com>
To: SPARQL Working Group <public-rdf-dawg@w3.org>
Spotted by the engineering team here.

http://www.w3.org/TR/2012/WD-sparql11-query-20120724/#func-strafter

This example seems to be wrong:
strafter("abc"^^xsd:string,"")	""^^xsd:string
I believe it should be 
strafter("abc"^^xsd:string,"")	"abc"^^xsd:string

The fn:substring-after says "If the value of $arg2 is the zero-length string, then the function returns the value of $arg1."

- Steve

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Received on Monday, 5 November 2012 12:34:39 GMT

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