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[XQuery Talk Mailing List Archive Home] [By Date] [By Thread] [By Subject] [By Author] [Recent Entries] [Reply To This Message] the least stupid way to just print the current date in xqueryGeert Josten geert.josten at dayon.nlFri Dec 2 00:02:24 PST 2011
I think the most elegant way would be something like: let $today := current-date() let $year := year-from-date($today) let $month := month-from-date($today) let $day := day-from-date($today) return concat($year, '-', $month, '-', $day) Kind regards, Geert -----Oorspronkelijk bericht----- Van: http://x-query.com/mailman/listinfo/talk [mailto:http://x-query.com/mailman/listinfo/talk] Namens http://x-query.com/mailman/listinfo/talk Verzonden: vrijdag 2 december 2011 8:52 Aan: http://x-query.com/mailman/listinfo/talk Onderwerp: the least stupid way to just print the current date in xquery Gentlemen, its my "virgin" post to the list. Allow me to test the waters by asking what's the least stupid way to just print the current date in xquery? (Without the ugly timezone baggage.) $ basex -q 'tokenize(string(current-date()),"\+")[1]' 2011-12-02 gives me exactly what I want, but it is rather 'expensive'. Other solutions involving fn:substring might break in 10000-2011=7989 years. Yes, something like the opposite of http://www.w3.org/TR/xpath-functions/#func-timezone-from-date or something(s) that would NOT just give me 2011-12-2 Hmm, fn:substring-before might break if I move to the other side of Greenwich, as will my tokenize without [+-] erg... Enough. How do you guys just print the current date? _______________________________________________ http://x-query.com/mailman/listinfo/talk http://x-query.com/mailman/listinfo/talk
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