From 09063ddab885bd432d09d0d2205511edbf6fe2f9 Mon Sep 17 00:00:00 2001 From: Eike Rathke Date: Wed, 1 Jun 2016 12:59:37 +0200 Subject: eliminate the mentioning of locale dependent date string in ISLEAPYEAR Rather use DATE() and "yyyy-mm-dd" ISO examples. Change-Id: I41c9dc57167f8c70e8674fcd4b4038ab8e835141 (cherry picked from commit 8353d5baa0186d0774a9ddd598677b31bff5922c) Reviewed-on: https://gerrit.libreoffice.org/25770 Reviewed-by: Eike Rathke Tested-by: Eike Rathke --- source/text/scalc/01/04060111.xhp | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) (limited to 'source/text') diff --git a/source/text/scalc/01/04060111.xhp b/source/text/scalc/01/04060111.xhp index 89de050211..7f6950cd1c 100644 --- a/source/text/scalc/01/04060111.xhp +++ b/source/text/scalc/01/04060111.xhp @@ -49,12 +49,12 @@ ISLEAPYEAR Determines whether a year is a leap year. If yes, the function will return the value 1 (TRUE); if not, it will return 0 (FALSE). Syntax -ISLEAPYEAR("Date") +ISLEAPYEAR(Date) -Date specifies whether a given date falls within a leap year. The Date parameter must be a valid date according to the locale settings of %PRODUCTNAME. +Date specifies whether a given date falls within a leap year. The Date parameter must be a valid date. Example =ISLEAPYEAR(A1) returns 1, if A1 contains 1968-02-29, the valid date 29th of February 1968 in your locale setting. -You may also use =ISLEAPYEAR("1968-02-29") or =ISLEAPYEAR("2/29/68"). +You may also use =ISLEAPYEAR(DATE(1968;2;29)) or =ISLEAPYEAR("1968-02-29") giving the date string in the ISO 8601 notation. Never use =ISLEAPYEAR(2/29/68), because this would first evaluate 2 divided by 29 divided by 68, and then calculate the ISLEAPYEAR function from this small number as a serial date number.
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