How PHP converts dates to JDE's Julian calendar format

Source: Internet
Author: User
This article mainly introduces the conversion of the date of PHP to JDE of the Julian calendar format method, has a certain reference value, now share to everyone, the need for friends can refer to

JDE's Julian calendar format rules are as follows:

The Julian calendar is a 6-digit number; (Example: January 1, 2018 = "118001)

The first digit represents the Century (example: 1 represents 21st century; 0 represents 20th century);

The 23rd digit represents the Year (example: 2018 is 18);

The number of days after the first 3 digits of the 1 year;

Here's how:

/** * Date converted to Jde Calendar "Support time Conversion Range: 1970-2,999" * "Six-bit: first identity century Example: 0 for 20th century, 1 for 21st century, second to third for year, and the last three for the first day of the Year" * Example: 2018-01-01  = "118001 * @param $date * @return string */function getjdedate ($date) {    #转换为时间戳    $unix _time = Strtotime ($date); 
  
    #获取时间信息    $ary _date = getdate ($unix _time);    #获取年    $str _year = $ary _date[' year ');    #获取世纪标识 "20th century = 0; 21st century = 1; 22nd century = 2 "    #如果年/100 Surplus    if ($str _year%100) {            $century = ceil ($str _year/100)%10; #向上取整    } else {            $century = Floor ($str _year/100)%10; #向下取整        }    #获取年后两位    $year = substr ($str _year,2);    #获取一年中的第几天    $year _day = $ary _date[' yday ') + 1;    #如果不足三位数补足三位数    $year _day = Str_pad ($year _day,3,0,str_pad_left);    #儒日历    return $century. $year. $year _day;}
  

Test Case:

<?php/** * Date converted to Jde Calendar "Support time Conversion Range: 1970-2,999" * "Six-bit: first identity century Example: 0 for 20th century, 1 for 21st century, second to third for year, and the last three for the first day of the Year" * Example: 2018-01-  118001 * @param $date * @return string */function getjdedate ($date) {    #转换为时间戳    $unix _time = Strtotime ($da TE);    #获取时间信息    $ary _date = getdate ($unix _time);    #获取年    $str _year = $ary _date[' year ');    #获取世纪标识 "20th century = 0; 21st century = 1; 22nd century = 2 "    #如果年/100 Surplus    if ($str _year%100) {        $century = ceil ($str _year/100)%10; #向上取整    } else {        $century = Floor ($str _year/100)%10; #向下取整    }    #获取年后两位    $year = substr ($str _year,2);    #获取一年中的第几天    $year _day = $ary _date[' yday ') + 1;    #如果不足三位数补足三位数    $year _day = Str_pad ($year _day,3,0,str_pad_left);    #儒日历    return $century. $year. $year _day;} echo getjdedate (' 2018-01-01 00:00:00 ');

Output:

118001

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