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  • JDToGregorian()

    (PHP 4, PHP 5, PHP 7)

    转变一个Julian Day计数为Gregorian历法日期

    说明

    jdtogregorian(int $julianday): string

    转变一个julian天数为gregorian历法的“月/日/年”形式的日期

    参数

    $julianday

    一个julian天数

    返回值

    “月/日/年”形式的gregorian日期

    参见

    • gregoriantojd() 转变一个Gregorian历法日期到Julian Day计数
    • cal_from_jd()转换Julian Day计数到一个支持的历法。
    JD days may have decimal fractions which correspond to the time of day. The Julian day begins at noon, and the decimal fraction measures fractional days until the beginning of the next day at noon.
    For instance, Julian Day 2453179.00000 is June 22, 2004, at 12:00pm (noon). 
    One hour later, it's 2453179.04167
    At 2453179.20833 I'll have dinner, and
    at 2453179.45833, it's time for the evening news.
    After a good night of sleep, my alarm will go off at 2453179.83333, 
    then at noon on June 23, a new Julian Day begins at 2453180.
    To use these functions with fractional days, strip the fractional part with floor(), and apply the function to the integer part. 
    Then add 12 hours, bringing you to noon of that day. That is the actual time returned by JDToGregorian().
    Then add the fractional part of the day, by multiplying the decimal part of the Julian Day by (24*60*60) seconds. This may take you forward or backward to a different Gregorian calendar date.
    Julian to Gregorian date change.
    If you do not have the calendar extensions loaded this is little function works realy well.
    <?php
    function jd_to_greg($julian) {
      $julian = $julian - 1721119;
      $calc1 = 4 * $julian - 1;
      $year = floor($calc1 / 146097);
      $julian = floor($calc1 - 146097 * $year);
      $day = floor($julian / 4);
      $calc2 = 4 * $day + 3;
      $julian = floor($calc2 / 1461);
      $day = $calc2 - 1461 * $julian;
      $day = floor(($day + 4) / 4);
      $calc3 = 5 * $day - 3;
      $month = floor($calc3 / 153);
      $day = $calc3 - 153 * $month;
      $day = floor(($day + 5) / 5);
      $year = 100 * $year + $julian;
      if ($month < 10) {
        $month = $month + 3;
      }
      else {
        $month = $month - 9;
        $year = $year + 1;
      }
      return "$day.$month.$year";
    }
    ?>
    
    The minimum acceptable input Julian day count is 1, which produces an output of "11/25/-4714" (at least in my operating system and location) which means 25 November 4714 BC. PHP does not recognize the year 0. Astronomers do use the year 0, and would write the Gregorian date that corresponds to Julian day number 1 as 25 November -4713.
    The php gregoriantojd() and jdtogregorian() functions, in addition to the limitations noted by httpwebwitch, does not take account of the 'Astronomical' system of reckoning - i.e. using a year zero, instead of going directly from 1BCE to 1CE, as with the Christian Anno Domini system.
    These functions can be used to wrap the php built-ins to return ISO 8601 compliant dates:
    <?php
    function ISO8601toJD($ceDate) {
      list($day, $month, $year) = array_map('strrev',explode('-', strrev($ceDate), 3));
      if ($year <= 0) $year--;
      return gregoriantojd($month, $day, $year);
    }
    function JDtoISO8601($JD) {
      if ($JD <= 1721425) $JD += 365;
      list($month, $day, $year) = explode('/', jdtogregorian($JD));
      return sprintf('%+05d-%02d-%02d', $year, $month, $day);
    }
    ?>
    
    I have made a slight modification to treebe's jd to greg function, this one will transform a unix timestamp to Gregorian day/month/year format...
    <?php
    function unix_to_greg($unix_timestamp) {
      $julian = floor(((($unix_timestamp / "60") / "60") / "24") + "2440588");
      $julian = $julian - 1721119;
      $calc1 = 4 * $julian - 1;
      $year = floor($calc1 / 146097);
      $julian = floor($calc1 - 146097 * $year);
      $day = floor($julian / 4);
      $calc2 = 4 * $day + 3;
      $julian = floor($calc2 / 1461);
      $day = $calc2 - 1461 * $julian;
      $day = floor(($day + 4) / 4);
      $calc3 = 5 * $day - 3;
      $month = floor($calc3 / 153);
      $day = $calc3 - 153 * $month;
      $day = floor(($day + 5) / 5);
      $year = 100 * $year + $julian;
      
      if ($month < 10)
      {
        $month = $month + 3;
      }else{
        $month = $month - 9;
        $year = $year + 1;
      }
      return "$day.$month.$year";
    }
    ?>
    

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