Question

我正在尝试向用户显示用户友好的日期格式，例如“1小时15分钟”，“4天8小时”。但是由于某种原因，我的脚本显示0小时为23小时。

 $date = '2014-01-15 15:00' # PAST DATE
    $now        = Date("Y-m-d H:m:s");
            $seconds    = strtotime($now) - strtotime($date);

            $days    = floor($seconds / 86400);
            $hours   = floor(($seconds - ($days * 86400)) / 3600);
            $minutes = floor(($seconds - ($days * 86400) - ($hours * 3600))/60);
            $seconds = floor(($seconds - ($days * 86400) - ($hours * 3600) - ($minutes*60)));

            if($days > 0)
            {
                if($days == 1)
                {
                    return $days . ' dag ' . $hours . ' timmar';
                } else {
                    return $days . ' dagar ' . $hours . ' timmar';
                }
            }

            if(($hours < 24) AND ($hours > 0))
            {
                return $hours . ' timmar';
            }

            if($minutes < 60)
            {
                return $minutes . ' minuter';
            }

任何人都可以看到导致这种情况的原因吗？我这样做是正确的吗？请注意，$ date是过去用户提供的。

Answer 1

有更简单的方法可以做到这一点：

$past = new DateTime('2014-01-15 15:00');
$now = new DateTime();
$interval = $now->diff($past);
echo $interval->format('%y years, %m months, %d days, 
                                %h hours, %i minutes, %S seconds');

显而易见的改进是现在使用显示零值（即0天）的时间段：

$elapsed = $interval->format('%y years, %m months, %d days, 
                              %h hours, %i minutes');
$elapsed = str_replace(array('0 years,', ' 0 months,', ' 0 days,', 
                            ' 0 hours,', ' 0 minutes,'), '', $elapsed);
$elapsed = str_replace(array('1 years, ', ' 1 months, ', ' 1 days, ', 
                             ' 1 hours, ', ' 1 minutes'), array('1 year, ', 
                             '1 month, ', ' 1 day, ', ' 1 hour, ', ' 1 minute'), 
                       $elapsed);

Answer 2

使用类似的东西：

            $date = '2014-01-15 15:00' # PAST DATE
            now        = Date("Y-m-d H:m:s");
            $t    = strtotime($now) - strtotime($date);    

            $time = date('g:iA M dS', $t );
            $diff = time() - $t;
            if ( $diff < 60 )
            {
                return "a few seconds ago";
            }
            elseif ( $diff < 3600 )
            {
                return "about ".( int ) ($diff/60) ." mins ago";
            }
            elseif ( $diff < 86400 )
            {
                if ( ( int ) ($diff/3600) == 1 )
                {
                    return "about an hour ago";
                }
                else
                {
                    return "about ".( int ) ($diff/3600) ." hours ago";
                }
            }
            else if( $diff < 172800 )
            {
                return "about a day ago";
            }
            elseif ( $diff > 172800 )
            {
                return "$time"; 
            }

用户友好的日期格式

2 个答案: