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我需要将日期与今天进行比较。基本上我输入了 2/17/13,它会输出“昨天”。我已经尝试过echo date('l jS F', strtotime('2/15/13'));,但它只会显示该日期而不是与今天进行比较。

4

5 回答 5

0

来自http://www.highlystructured.com/comparing_dates_php.html

$exp_date = "2006-01-16";
$todays_date = date("Y-m-d");

$today = strtotime($todays_date);
$expiration_date = strtotime($exp_date);

if ($expiration_date > $today) {
     $valid = "yes";
} else {
     $valid = "no";
}
于 2013-02-18T16:10:57.987 回答
0

这是一个示例,可以帮助您入门:

$when = '2/17/13';
$now  = new DateTime();
// I append $now's time to $then, to make sure we compare 
// the two dates using the same time of the day
// if you want to compare to $then at 00:00:00, leave it out
$then = new DateTime( $when . ' ' . $now->format( 'H:i:s' ) );

$diff = $now->diff( $then );
switch( $diff->days )
{
    case 0:
        echo 'today';
    break;
    case 1:
        echo $diff->invert ? 'yesterday' : 'tomorrow';
    break;
    default:
        echo $diff->invert ? $diff->days . ' days ago' : $diff->days . ' days from now';
    break;
}
于 2013-02-18T16:23:56.377 回答
0

Drupal 附带了一个非常好的format_interval($interval, $granularity = 2, $langcode = NULL) 功能

<?php
function format_interval($interval, $granularity = 2, $langcode = NULL) {
  $units = array(
    '1 year|@count years' => 31536000,
    '1 month|@count months' => 2592000,
    '1 week|@count weeks' => 604800,
    '1 day|@count days' => 86400,
    '1 hour|@count hours' => 3600,
    '1 min|@count min' => 60,
    '1 sec|@count sec' => 1,
  );
  $output = '';
  foreach ($units as $key => $value) {
    $key = explode('|', $key);
    if ($interval >= $value) {
      $output .= ($output ? ' ' : '') . format_plural(floor($interval / $value), $key[0], $key[1], array(), array('langcode' => $langcode));
      $interval %= $value;
      $granularity--;
    }

    if ($granularity == 0) {
      break;
    }
  }
  return $output ? $output : t('0 sec', array(), array('langcode' => $langcode));
}
?>

您无需运行 Drupal 即可使用它。只需在某处包含上述功能。但是,上面的函数也调用了format_plural()- 调用t()(两个自定义 Drupal 函数),所以你需要修改上面的函数或者包含所有的函数。

您的用例:

<?php
$today = date();
$compare = strtotime('2/15/13');
$interval = ($today - $compare);

print format_interval($interval, 1);
// Outputs '1 day'
?>
于 2013-02-18T16:29:32.303 回答
0
function checkDate($todaysDate,$expirationDate){
 $today = strtotime($todays_date);
 $expiration_date = strtotime($exp_date);

 if ($expiration_date > $today) {
     $valid = "yes";
 } else {
     $valid = "no";
 }

return $valid
}


$expirationDate = "2006-01-16";
$todaysDate = date("Y-m-d");

$valid=checkDate($todaysDate,$expirationDate)

在函数中执行此操作,然后您可以将其与任何日期组合一起使用。然后,您可以使用一些条件扩展该函数,以确定它是否“昨天”。

于 2013-02-18T16:34:23.357 回答
0
function relativeTime($time) {

$d[0] = array(1,"second");
$d[1] = array(60,"minute");
$d[2] = array(3600,"hour");
$d[3] = array(86400,"day");
$d[4] = array(604800,"week");
$d[5] = array(2592000,"month");
$d[6] = array(31104000,"year");

$w = array();

$return = "";
$now = time();
$diff = ($now-$time);
$secondsLeft = $diff;

for($i=6;$i>-1;$i--)
{
     $w[$i] = intval($secondsLeft/$d[$i][0]);
     $secondsLeft -= ($w[$i]*$d[$i][0]);
     if($w[$i]!=0)
     {
        $return.= abs($w[$i]) . " " . $d[$i][1] . (($w[$i]>1)?'s':'') ." ";
     }

}

$return .= ($diff>0)?"ago":"left";
return $return;
}
于 2013-02-18T16:42:54.887 回答