PHP 怎么列出来某一个日期距离现在的日期

Question

比如
$aa="2013-12-25";
$bb=time();
我想列出2013-12距离现在的月份

array(12) {
[1] => string(6) "2013-12"
[2] => string(6) "2014-1"
[3] => string(6) "2014-2"
[4] => string(6) "2014-3"
[5] => string(6) "2014-...."
}

现在只能列出当年的当前月份
for ($i = 1; $i <= date('m'); $i++) {
$dates[$i] = date('Y-'.$i, strtotime(date("Y-m-d")));
}

Answer 1

先说感想吧

• 这种“要知道循环多少次略有困难”的问题用while会让逻辑清晰不少
• strtotime大杀器
• PHP4.3+
• //我的代码又短又清楚，哼

php//http://3v4l.org/vuef4
function monthToToday($past, $now, $format = 'Y-m') {
    $current = strtotime(date('Y-m-1', strtotime($past)));
    $result = array();
    while($current < $now) {
        $result[] = date($format, $current);
        $current = strtotime('+1 month', $current);
    }

    //包含当前月份
    $result[] = date($format, $current);

    return $result;
}

print_r(monthToToday('2013-12-25', time()));
die;

Answer 2

echo (new DateTime('2013-12'))->diff(new DateTime(date('Y-m')))->format('%y year %m month %d days');

结果是从目标时间到今天已经过去几年几月几天

年x12+month = 已经过去几个月

$date = new DateTime('2013-12');
$now = date('Ym');
$months = array();
while ($date->format('Ym')<=$now) {
    array_push($months, $date->format('Y-m'));
    $date->modify('+1 month');
}
print_r($months);

获取目标时间到现在的月份详情

Answer 3

两者时间戳相减

Answer 4

<?php
// http://3v4l.org/HkTl8
function diffMonth($month, $format = "Y-m", $monthDiffed = "now", $dateTimeZone = false) {
    if(!$dateTimeZone) $dateTimeZone = new DateTimeZone("Asia/Shanghai");

    $monthDiffed = new DateTime($monthDiffed, $dateTimeZone);
    $month = new DateTime($month, $dateTimeZone);
    if($month > $monthDiffed) {
        $m = $month;
        $month = $monthDiffed;
        $monthDiffed = $m;
    }
    $monthDiffed = $monthDiffed->modify("last day of -1 month");
    $month = $month->modify("first day of +1 month");

    $result = array();
    while($month <= $monthDiffed) {
        $result[] = $month->format( $format );
        $month = $month->modify("+1 month");
    }
    return $result;
}

print_r( diffMonth("2014-6") );
print_r( diffMonth("2014-6", "Y/m", "2013-6") );

Answer 5

$aa="2003-02-25";

$bb = date("Y-m-d",time());


function datediffage($before, $after) {

    $b = getdate($before);
    $a = getdate($after);


$y=$m=0;

if ($a['mon']==1) { //1月，借年
    $y=$a['year']-$b['year']-1;$m=$a['mon']-$b['mon']+12;
}
else {
    if ($a['mon']>=$b['mon']+1) { //借月后，月相减为正
        $y=$a['year']-$b['year'];$m=$a['mon']-$b['mon']-1;
    }
    else { //借月后，月相减为负，借年
        $y=$a['year']-$b['year']-1;$m=$a['mon']-$b['mon']+12-1;
    }
}

$datearr = array();

$totalm = "";
if( $y > 0 ){
    $totalm = $y*12;//大于1年，转换成月份数
}

$totalm += $m;

for ($i=0; $i < $totalm; $i++) { 
    array_push($datearr, date("Y-m",mktime(0,0,0,$b['mon']+$i,$d,$b['year'])));
}

return $datearr;

}

print_r(datediffage(strtotime($aa),strtotime($bb)));

圆代码来自：http://my.oschina.net/u/223350/blog/293687
感谢原作者，有修改。

Answer 6

$aa = '2013-12-25';
$a = strtotime($aa);
$b = array();
while($a<time()){
$b[] = date('Y-m',$a);
$a = strtotime('+1 month',$a);
}
var_dump($b);

写完发现和mcfog思路差不多

Answer 7

这个问题貌似都有问题啊，
时间差出来了，那么，你一个月是按28天算，还是30天或31天算呢？
年是按365天还是366天算呢？