JDDayOfWeek
(PHP 4, PHP 5, PHP 7)
JDDayOfWeek — 返回星期的日期
说明
$julianday
[, int $mode = CAL_DOW_DAYNO
] )返回星期的日期，根据模式不同可能是字符串或是整数。
参数
-
julianday -
一个julian天数。
-
mode -
历法星期模式 Mode Meaning 0 (默认) 返回数字形式(0=Sunday, 1=Monday, etc) 1 返回字符串形式 (English-Gregorian) 2 返回缩写形式的字符串 (English-Gregorian)
返回值
数字或字符串形式的星期数。
用户评论:
[#1] alex a khimich d org [2015-09-27 19:01:23]
In my case i had lots of head pain with returning day number using this function.
Here is bypass solution using just date():
<?php
$first_day_of_September=date('w', strtotime('1-09-2015')); // 2
?>
[#2] nrkkalyan at rediffmail dot com [2005-02-25 22:08:11]
You can get todays day time and date using this code
<?php
echo date("d")." ";
echo date("m")." ";
echo date("Y")." ";
echo date("h:i:s A");
ECHO ' <br/>';
echo jddayofweek ( cal_to_jd(CAL_GREGORIAN, date("m"),date("d"), date("Y")) , 1 );
?>
[#3] php at xtramicro dot com [2004-09-07 08:28:57]
Be aware that date() and mktime() only work as long as you move within the UNIX era (1970 - 2038 / 0x0 - 0x7FFFFFFF in seconds). Outside that era those functions are only generating errors.
In other words: mktime(0, 0, 0, 12, 31, 1969) *DOES NOT* work (and so doesn't date() fed with with mktime()'s result from above). But cal_to_jd(CAL_GREGORIAN, 12, 11, 1969) *DOES WORK*.
And please note that the calendar-extension's functions arguments follow the US date order: month - day - year.