Question

我正在从Wordpress字段中获取日期，我需要检查日期是否已过去或仍未来。

    $dates = ['date'=>'02/12/13','date'=>'10/12/14','date'=>'14/01/15'];


    foreach ($dates as $date){

        $the_date = $date['date'];


        echo $the_date;

        echo "  ";

        echo date('d/m/y');

        echo "  ";

        if($the_date < date('d/m/y')){
            echo 'gone';
        }else{
            echo 'to come';
        }

    }

foreach回应了这一点。

    02/12/13 22/11/14 gone

    10/12/14 22/11/14 gone

    14/01/15 22/11/14 gone

    27/01/15 22/11/14 to come

    10/02/15 22/11/14 gone

看起来它只是检查第一天的日期。

Answer 1

更好的选择是使用DateTime类。它允许使用comparison operators比较两个DateTime实例。

$dates = ['02/12/13','10/12/14','14/01/15'];

foreach ($dates as $date) {
    $the_date = \DateTime::createFromFormat('d/m/y', $date);
    $now = new \DateTime();

    echo $date." ".($the_date < $now ? 'gone' : 'to come')."\n";
}

您看到的问题是因为日期被比较为字符串。目前的日期是＆＃34; 22/11/14＆＃34;因此它将更大比任何其他日期开始的日期为＆＃34; 1＆＃34;或＆＃34; 0＆＃34;。

PD：您的数组包含许多使用相同＆＃39;日期＆＃39;键。这是一个问题所以我在我的例子中删除了它们。

Answer 2

<?php

$dates = array('02/12/13','10/12/14','14/01/15');

$now = mktime(0,0,0);
foreach($dates as $date) {
  $tmp = explode('/',$date);
  $date_time = mktime(0,0,0,intval($tmp[1]),intval($tmp[0]),intval($tmp[2]));
  echo $date . ' ' . ($now > $date_time?'gone':'to come') . "\n";
}

Answer 3

使用PHP的DateTime API：

$date='02/12/13';

if(\DateTime::createFromFormat('d/m/y',$date) < new \DateTime()){
//date is in the past
}else{
//date is either today or in the future
}

官方PHP文档：

http://php.net/manual/en/class.datetime.php

Answer 4

最好的方法是使用时间戳：试试这个：

foreach ($dates as $date){

    $the_date = $date['date'];


    echo $the_date;

    echo "  ";

    echo date('d/m/y');

    echo "  ";

    if( strtotime($the_date) < time() )
    {
        echo ' is gone';
    }
    else
    {
        echo ' is to come';
    }

}

Answer 5

保持简单：

// $date is the date you need to compare to today
$date = ("2015 10 03");

// Make sure their formats are purely numeric and match
if ($date->format('m.d.y') >= date('m.d.y'))
{
   your procedure...
}

Answer 6

我建议使用DateTime类的功能。然后你可以按如下方式进行检查：

<?php
$then = $reset_date;
$then = new DateTime($then);
$now = new DateTime(date("m-d-Y"));
$sinceThen = $then->diff($now);
$new = new DateTime($reset_date);
$old = new DateTime(date("m-d-Y"));

if ( $old->modify('+1 year') < $new) {
    echo "<font color='red'>Reset now <br></font>";
    echo "<font color='orange'>$sinceThen->y years <br></font>";
    echo "<font color='orange'>$sinceThen->m months </font>";
    echo "<font color='orange'>$sinceThen->d days have passed.<br></font>";
} else {
    echo "<font color='green'> $sinceThen->y years <br>
          $sinceThen->m months  $sinceThen->d days till to Reset.</font>";
    //Combined
 }
?>

php检查日期是否已过

6 个答案: