需要修改日期倒计时的PHP代码（当计数达到0时不显示任何内容）

需要修改日期倒计时的PHP代码（当计数达到0时不显示任何内容）

我使用下面的代码来显示一个特定日期的文本视觉倒计时，分为月、周、日。效果很好。但是，当倒计时到达实际事件日期，然后超过事件日期时，它仍然显示绝对差异。我希望代码此时不显示任何内容（不向屏幕输出）。我希望能就如何修改这一点提供一些指导 要按说明执行的代码

     $d1 = new DateTime();  // now
     $d2 = new DateTime('2014-01-08');  // set the date +1 to compensate for 1-day  
      error in script
     $diff = $d2->diff($d1);
     list($y,$m,$d) = explode('-', $diff->format('%y-%m-%d'));
     $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':'');}

试试这个：

<?php

$d1 = new DateTime();  // now
$d2 = new DateTime('2014-01-08');  // set the date +1 to compensate for 1-day  
$diff = $d2->diff($d1);

list($y,$m,$d) = explode('-', $diff->format('%y-%m-%d'));
if ($d1 < $d2) {
    $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':'');}
}

---

您需要检查差异是正的还是负的。由于
diff
方法返回
DateInterval
，因此可以检查
invert
属性

 $d1 = new DateTime();  // now
 $d2 = new DateTime('2014-01-08');

 $diff = $d2->diff($d1);
 if ($diff->invert == 1) // the countdown is running
 {
     list($y,$m,$d) = explode('-', $diff->format('%y-%m-%d'));
     $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 
 {
     // The countdown finished, do something!
 }

有很多方法可以做到这一点

我认为比较容易的是：

$FinalDate='2013-09-23';
$d1 = new DateTime();  // now     
$d2 = new DateTime($FinalDate);  // set the date +1 to compensate for 1-day  

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

if(strtotime($FinalDate)>time()){
 printf('Countdown To Event - ');
 if ($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 "Actions AFTER the date";
 }

然而，这段代码看起来非常糟糕，您可以做得更好。
你可以用javascript来做，还可以放一个漂亮的秒计数器
 谢谢。这是一个有效的选择。现在我将使用更简单的if（$d1<$d2）{}建议。谢谢。这是一个有效的选择。现在我将使用更简单的if（$d1<$d2）{}建议。