Hello, Let's try to use the SPARQL grammar rule [80] STRING_LITERAL_LONG1 ::= "'''" ( [^'\] | ECHAR | ("'" [^']) | ("''" [^']) )* "'''" to parse a string '''some-plain-text''\\more-plain-text''' It seemes to me that it is parsed as -- starting "'''"; -- "some-plain-text", in a char-by-char loop, -- "''\", matching ("''" [^']), -- "\m", oops, this does not match ECHAR. -- ore-plain-text''' is parsed by error recovery code. I believe that the proper formal notation is [80] STRING_LITERAL_LONG1 ::= "'''" ( ("'" ("'"?))? ([^'\] | ECHAR) )* "'''" Unfortunately, this is less readable. Best Regards, Ivan Mikhailov OpenLink Software.Received on Friday, 30 December 2005 08:14:04 GMT
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