Namejddayofweek — Gets the day of the week for the Gregorian calendar from the given Julian day count.
jddayofweek() finds the day of the week for the Gregorian calendar date that corresponds to the specified Julian day count. The day of the week is returned as an integer or as the full or abbreviated name of the day.
The flag parameter controls the format in which the day of the week is output. If the flag argument is not set, flag is 0. The flags are described in the following table.
You may be wondering, if the default value for flag is 0, how the caller can tell the difference between an error and Sunday. The user uses the strict comparison operator (===); however, the function doesn't return FALSE on error. This function is poorly designed and only indicates that an error has occurred if the user doesn't give the function an argument. In this case, NULL is returned and a warning is generated.
To get a Julian day count from another calendar system, see the various *tojd() functions.
$year = date ('Y'); // Find the current year $xmas = gregoriantojd (12, 25, $year); $new_years = gregoriantojd (1, 1, $year + 1); echo 'Christmas is on a ', jddayofweek( $xmas, 1); echo ' and New Year\'s Day is on a ', jddayofweek( $new_years, 1) . '.';
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