Php - 制定出什么学年

问题描述

学年每年8月1日开始。

我如何计算出目前日期的学年。

ie
2012年7月31日（20120631） ）将作为2011/2012计划。

2012年8月13日（20120801）将作为2012/2013计划。

在这一刻，这是我所拥有的，但它不是很好，因为我不想定义日期，它不会返回正确的学年只是原来定义的$ academic_start_date。

  function check_in_range（$ start_date，$ end_date，$ date_from_user）
 {
 //转换为时间戳
 $ start_ts = strtotime（$ start_date ）; 
 $ end_ts = strtotime（$ end_date）; 
 $ user_ts = strtotime（$ date_from_user）; 
 
 //检查用户日期是否在start& end 
 return（（$ user_ts> = $ start_ts）&（$ user_ts< = $ end_ts））; 
} 
 
 
 $ academic_start_date ='20110801'; 
 $ academic_end_date ='20120731'; 
 $ startdate ='20120813'; 
 
 $ acyear_check = check_in_range（$ academic_start_date，$ academic_end_date，$ startdate）; 
 if（$ acyear_check == 1）{$ acyear = $ academic_start_date;} 
 else {$ acyear ='';} 
  

解决方案

  $ time = ??; //这里你放时间戳，它也可能是strtotime smth）; 
 
 $ year = date（'Y'，$ time）; 
 if（date（'n'，$ time）< 8）
 $ ayear =（$ year  -  1）。 
 else 
 $ ayear =（$ year）。'/'。（$ year + 1）; 
  

对于您的格式：

  $ datestr ='YYYYmmdd'; 
 $ year = substr（$ datestr，0，4）; 
 if（intval（substr（$ datestr，4,2））< 8）
 $ ayear =（$ year  -  1）。 
 else 
 $ ayear =（$ year）。'/'。（$ year + 1）; 
  

I am trying to write a script that would work out the current academic year.

Academic year starts on Aug 1st every year.

How can i work out which Academic year we are in based on the current date.

ie31st July 2012 (20120631) would work out as 2011/2012

13th Aug 2012 (20120801) would work out as 2012/2013

At the moment this is what I have, but its not very good as i dont want to define the dates and it does not return the correct academic year just the originally defined$academic_start_date.

function check_in_range($start_date, $end_date, $date_from_user)
{
  // Convert to timestamp
  $start_ts = strtotime($start_date);
  $end_ts = strtotime($end_date);
  $user_ts = strtotime($date_from_user);

  // Check that user date is between start & end
  return (($user_ts >= $start_ts) && ($user_ts <= $end_ts));
}


$academic_start_date = '20110801';
$academic_end_date = '20120731';
$startdate = '20120813';

$acyear_check = check_in_range($academic_start_date, $academic_end_date, $startdate);
if ($acyear_check == 1) { $acyear = $academic_start_date;}
else { $acyear = '';}

解决方案

$time = ??;// here you put timestamp, it also may be strtotime(smth);

$year = date('Y', $time);
if(date('n', $time) < 8)
     $ayear = ($year - 1).'/'.$year;
else
    $ayear = ($year).'/'.($year + 1);

For your format:

$datestr = 'YYYYmmdd';
$year = substr($datestr, 0, 4);
if(intval(substr($datestr,4,2)) < 8)
     $ayear = ($year - 1).'/'.$year;
else
    $ayear = ($year).'/'.($year + 1);

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