Question

嗨我想显示今天，如果给定的日期是今天显示倒计时..

这是我到目前为止所尝试过的..它总是什么都不显示......我做错了什么？

$d1 = new DateTime();  // now
$d2 = new DateTime('2018-01-07');  // set the date +1 to compensate for 1-day  
$diff = $d2->diff($d1);


list($y,$m,$d) = explode('-', $diff->format('%y-%m-%d'));
if ($d1 < $d2) {
    $months = $y*12 + $m;
    $weeks = floor($d/7);
    $days = $d%7;

     if($diff==0)
 {
    echo 'today';
 }
    else
    {
    printf('Countdown To Event : ');
    if ($months) {printf('%d month%s ', $months, $months>1?'s':'');}
    if ($weeks) {printf('%d week%s ', $weeks, $weeks>1?'s':'');}
    if ($days) {printf('%d day%s ', $days, $days>1?'s':'');}
    }


}

Answer 1

您错误地使用DateInteval对象

$d1 = new DateTime();  // now
$d2 = new DateTime('2018-01-07');  // set the date +1 to compensate for 1-day  

// Object of DateInterval class
$diff = $d2->diff($d1);
// Difference in days
$d = $diff->days;

if (! $d) {
    echo 'today';
}
else  {
    $months = $diff->m; 
    $days = $diff->d;
    printf('Countdown To Event : ');
    if ($months) {printf('%d month%s ', $months, $months>1?'s':'');}
//    if ($weeks) {printf('%d week%s ', $weeks, $weeks>1?'s':'');}
    if ($days) {printf('%d day%s ', $days, $days>1?'s':'');}
}

demo

Answer 2

这应该可以解决问题：

<?php

$today = new DateTime();
$date = (new DateTime())->add(new DateInterval('P1D'));

if ($today->format('Y-m-d') === $date->format('Y-m-d')) {
  echo 'TODAY';
} else {
  $d = $today->diff($date);
  $result = '';

  if ($d->years > 1) {
      $result .= $d->years.' Years | ';
  } else if ($d->years == 1) {
      $result .= '1 Year | ';
  } else {
      $result .= '0 Years | ';
  }

  if ($d->months > 1) {
      $result .= $d->months.' Months | ';
  } else if ($d->months == 1) {
      $result .= '1 Month | ';
  } else {
      $result .= '0 Months | ';
  }

  if ($d->days > 1) {
      $result .= $d->days.' Days';
  } else if ($d->days == 1) {
      $result .= '1 Day';
  } else {
      $result .= '0 Days';
  }

  echo $result;
}

?>

工作演示here。

Answer 3

<?php
 $d1 = new DateTime();  // now
 $d2 = new DateTime('2019-01-08');  // set the date +1 to compensate for 1-
 day
 $diff = $d2->diff($d1);


list($y,$m,$d) = explode('-', $diff->format('%y-%m-%d'));
if ($d>0) {
$months = $y*12 + $m;
$weeks = floor($d/7);
$days = $d%7;


printf('Countdown To Event : ');
if ($months) {printf('%d month%s ', $months, $months>1?'s':'');}
if ($weeks) {printf('%d week%s ', $weeks, $weeks>1?'s':'');}
if ($days) {printf('%d day%s ', $days, $days>1?'s':'');}
}
else
{
echo "today";
}

$ diff是一个对象......不正确$ diff == 0

Answer 4

这是一个简短而简单的代码，可能会解决您的问题。请尝试： -

  <?php
$d1 = date("Y-m-d");
$d2 = '2018-01-06';


if(strtotime($d1) == strtotime($d2) )
{
echo 'today';
}
else
{
printf('Countdown To Event : ');
}?>

今天在倒计时代码中显示

4 个答案: