如何将公历日转换为公历日期

Question

我在我的 table 中存储了公历日期和一些其他数据。这些天我通过以下 Erlang 命令获得：

{Date, _} = calendar:now_to_datetime(now()).
GDays = calendar:date_to_gregorian_days(Date).

我们以这个值为例：GDays = 736202.

我用 ErlyDTL 在 Cowboy 中建立了一个网站以供参考。现在我想以日期格式显示这些公历。 (2015 年 8 月 28 日).

我使用以下代码从我认为的列表中获取数据：

{% for item in list %}
  {{item.1}} <br/>
{% endfor %}

我尝试了以下命令 {{item.1|date:" D d M Y"}} 但出现错误：

Unexpected date parameter: 736202

现在如何将公历日期转换为 erlyDTL 或 javascript 中的日期时间？

提前致谢

Answer 1

根据此处的文档：http://www.erlang.org/doc/man/calendar.html

Erlang 以 0 月 1 日这一纪元为基础，并提供了一个示例：1970 年 1 月 1 日是 719528。所以纪元是第 0 天。

以下函数将 Date 对象转换为公历日期并返回。他们 return 根据文档中的单个示例更正值，但将 736202 转换为 2015 年 8 月 27 日而不是 8 月 28 日。也许您使用的是 UTC 而不是本地时间。不管怎样，我想这里已经足够你解决了。

/* @param {Date} [date] - Date object to be converted
** @returns {number} - whole days since 1 January 0 to d
**
** epoch date must set year separately as in many implementations
** new Date(0,0,1) returns 1 Jan 1900, not 1 Jan 0000
*/
function dateToGregorianDays(date) {

  // Create a Date object for 0000-Jan-01 (months are zero based)
  var epoch = new Date(0,0,1);

  // Set the epoch to year 0 as in the above some browsers will
  // create a date for 1900 not 0, even though 0 was passed in
  epoch.setFullYear(0);

  // Copy the passed in Date so it's not modified by next step
  var e = new Date(+date);

  // Set the time part of the copied date to 00:00:00, which
  // helps to calculate whole days
  e.setHours(0,0,0,0);

  // In mathematic operations, dates are converted to their time value
  // which is milliseconds, so get the difference in milliseconds between
  // the two dates and divide by milliseconds per day. Round to remove 
  // fractional parts caused occasionally over daylight saving boundaries
  // to get whole day count between the two dates and return it
  return Math.round((e - epoch)/8.64e7);
}

/* @param {number} [days] - Gregorian day number
** @returns {Date} - Based on whole days since 1 January 0
**
**      0 -> 1 Jan 0000
** 719528 -> 1 Jan 1970
*/
function gregorianDaysToDate(days) {

  // Create a date for 0000-Jan-01
  var epoch = new Date(0,0,1);
  epoch.setFullYear(0);

  // Add the number of days to the date
  // epoch.getDate() could be replaced by 1 since that's what
  // the date was set to just above
  epoch.setDate(epoch.getDate() + days);

  // Return the date
  return epoch;
}

/* Simple function to return a date string as dd-MMM-yyyy
** @param {Date} [date] - Date to format
** @returns {string} - formatted string for date
*/
function formatDateDMY(date) {
 
  // Month names
  var months = ['Jan','Feb','Mar','Apr','May','Jun',
                'Jul','Aug','Sep','Oct','Nov','Dec'];

  // Add leading zero to single digit days
  // Get the month name for the month (zero indexed, 0 is Jan)
  // Add leading zeros to years with less than 4 digits
  // Use '-' as separator
  return ('0' + date.getDate()).slice(-2) + '-' +
         months[date.getMonth()] + '-' +
         ('000' + date.getFullYear()).slice(-4);
}

// Create an alias for the above function to save typing
var fd = formatDateDMY;

// Gregorian days for 01-Jan-1970
document.write(dateToGregorianDays(new Date(1970,0,1)) + '<br>'); // 719528

// Gregorian calendar date for 719528 formatted as dd-MMM-yyyy
document.write(fd(gregorianDaysToDate(719528)) + '<br>')          // 01 Jan 1970

// Gregorian calendar date for 0 formatted as dd-MMM-yyyy
document.write(fd(gregorianDaysToDate(0)) + '<br>');              // 01 Jan 0000

// Gregorian calendar date for 736202 formatted as dd-MMM-yyyy
document.write(fd(gregorianDaysToDate(736202)) + '<br>');         // 27 Aug 2015

// Gregorian days for 28-Aug-2015
document.write(dateToGregorianDays(new Date(2015,7,28)));         // 736203

Answer 2

这是我需要的算法。我在 javascript 中以一种非常简单的方式做到了，现在已经足够好了，所以不要挑剔 ;)。不支持闰年

<html>
  <head>
    <script>
      function date(n){
        // Calculate the year.
        var year = Math.round(n / 365.2425 - 0.5);
        // Get the remainder of the year calculation.
        var dec = Math.round(((n * 10000) / 365.2425) - (year * 10000));
        // Multiply remainder with 365.2425.
        var day = Math.round((dec * 365.25) / 10000);
        // Search which month and day it is.
        if(day <= 30){
          return day.toString().concat(" januari ").concat(year);
        }else if(day <= 58){
          return (day - 30).toString().concat(" februari ").concat(year.toString());
        }else if(day <= 89){
          return (day - 58).toString().concat(" maart ").concat(year.toString());
        }else if(day <= 119){
          return (day - 89).toString().concat(" april ").concat(year.toString());
        }else if(day <= 150){
          return (day - 119).toString().concat(" mei ").concat(year.toString());
        }else if(day <= 180){
          return (day - 150).toString().concat(" juni ").concat(year.toString());
        }else if(day <= 211){
          return (day - 180).toString().concat(" juli ").concat(year.toString());
        }else if(day <= 242){
          return (day - 211).toString().concat(" augustus ").concat(year.toString());
        }else if(day <= 272){
          return (day - 242).toString().concat(" september ").concat(year.toString());
        }else if(day <= 303){
          return (day - 272).toString().concat(" oktober ").concat(year.toString());
        }else if(day <= 333){
          return (day - 303).toString().concat(" november ").concat(year.toString());
        }else if(day <= 364){
          return (day - 333).toString().concat(" december ").concat(year.toString());
        }else{
          return "error";
        }
      }
    </script>
  </head>
  <body>
    <script>
      document.write(date(736205));
    </script>
  </body>
</html>

感谢大家的帮助。

如何将公历日转换为公历日期

How to convert gregorian days to gregorian date

javascript

erlang

datetime

django-templates