Question

如何检查是生日是在本周，2周，还是月份我用下面的代码检查但是它返回了错误的计算。

public function CountDown($birthdate, $days=7)
{
    list($y,$d,$m) = explode('/',$birthdate);
    $today = time();
    $event = mktime(0,0,0,$m,$d,$y);
    $apart = $event - $today;
    if ($apart >= -86400)
    {
        $myevent = $event;
    }
    else
    {
          $myevent = mktime(09,0,0,$m,$d,$y);
    }
    $countdown = round(($myevent - $today)/86400);
    if ($countdown <= $days)
    {
          return true;
    }
    return false;
}

Answer 1

试试这个：

function CountDown($birthdate, $days=7)
{
    # create today DateTime object
    $td = new DateTime('today');
    # create birth DateTime object, from format Y/d/m
    $bd = DateTime::createFromFormat('!Y/d/m', $birthdate);
    # set current year to birthdate
    $bd->setDate($td->format('Y'), $bd->format('m'), $bd->format('d'));
    # if birthdate is still in the past, set it to new year
    if ($td > $bd) $bd->modify('+1 year');
    # calculate difference in days
    $countdown = $bd->diff($td)->days;
    # return true if day difference is within your range
    return $countdown <= $days;
}

demo

Answer 2

这对我有用

class Birthday{
    public function CountDown($birthdate, $days=7)
    {
        list($y,$d,$m) = explode('/',$birthdate);
        $today = time();
        $event = mktime(0,0,0,$m,$d,$y);
        $apart = $event - $today;
        if ($apart >= -86400)
        {
            $myevent = $event;
        }
        else
        {
              $myevent = mktime(09,0,0,$m,$d);
        }
        $countdown = round(($myevent - $today)/86400);
        if (($countdown <= $days))
        {
              return true;
        }
        return false;
    }
}

$bday = new Birthday;
$count = $bday->CountDown("1969/16/11"); //today is 2014/14/11

var_dump($count); //returns true.

我刚从$ myevent中的mktime（）中删除了年份。这改变了答案是准确的。另一种方式是将倒数计算成一个巨大的负数。

使用php检查一周内或一个月内即将到来的生日

2 个答案: