Question

我有这个功能来计算自某个日期以来经过的相对时间

function nicetime($date) {
    if(empty($date)) {
        return "No date provided";
    }

    $periods         = array("second", "minute", "hour", "day", "week", "month", "year", "decade");
    $lengths         = array("60","60","24","7","4.35","12","10");

    $now             = time();
    $unix_date         = strtotime($date);

       // check validity of date
    if(empty($unix_date)) {   
        return "Bad date";
    }

    // is it future date or past date
    if($now > $unix_date) {   
        $difference     = $now - $unix_date;
        $tense         = "ago";

    } else {
        $difference     = $unix_date - $now;
        $tense         = "from now";
    }

    for($j = 0; $difference >= $lengths[$j] && $j < count($lengths)-1; $j++) {
        $difference /= $lengths[$j];
    }

    $difference = round($difference);

    if($difference != 1) {
        $periods[$j].= "s";
    }

    return "$difference $periods[$j] {$tense}";
}

我将该函数的日期传递为“2011-01-16 12:30”，但是我收到了错误的日期，这意味着$unix_date为空，但是如果我死在函数中我得到{ {1}}传递给函数然后它返回一个（在它之前，下面是我调用方法的方式。

$date

Answer 1

echo nicetime("2011-01-16 12:30");

对我来说很好。我建议你确保

 $a['created_at']

告诉你它是什么东西。也许试试

$dtDate = $a['created_at'];
$strDate = date('Y:m:d G:i', $dtDate );
echo $strDate;
echo nicetime($strDate);

确保$ strDate是您认为的。

php相对时间问题

1 个答案: