Question

我有一个日历显示星期几的错误日期。更具体地说，它们比实际日期落后一天。

我做了很多研究，并使用了以下SO问题（php date('d') calculates same output for two consecutive days），但我不能让它在我的脚本上工作。

它目前正在显示10月26日的两次，从日期出现错误。

$firstday = mktime(0,0,0,$month,1,$year);
$day = date('Y-m-d',$firstday);
$fday = strtotime($day." last Sunday ",$firstday);
$currday_timestamp = mktime(0,0,0,date('m'),date('d'),date('Y'));

for($i=0;$i<7;$i++) {
            $firstweek[$i]['id'] = date("Y-m-d",$fday+(86400 * $i));
            $firstweek[$i]['val'] = date("d",$fday+(86400 * $i));
            if($currday_timestamp > $fday+(86400 * $i)) {
                $firstweek[$i]['flag'] = 1;    
            }
            $secondweek[$i]['id'] = date("Y-m-d",$fday+(86400 * ($i+7)));
            $secondweek[$i]['val'] = date("d",$fday+(86400 * ($i+7)));
            if($currday_timestamp > $fday+(86400 * ($i+7))) {
                $secondweek[$i]['flag'] = 1;    
            }
            $thirdweek[$i]['id'] = date("Y-m-d",$fday+(86400 * ($i+14)));
            $thirdweek[$i]['val'] = date("d",$fday+(86400 * ($i+14)));
            if($currday_timestamp > $fday+(86400 * ($i+14))) {
                $thirdweek[$i]['flag'] = 1;    
            }
            $fourthweek[$i]['id'] = date("Y-m-d",$fday+(86400 * ($i+21)));
            $fourthweek[$i]['val'] = date("d",$fday+(86400 * ($i+21)));
            if($currday_timestamp > $fday+(86400 * ($i+21))) {
                $fourthweek[$i]['flag'] = 1;    
            }
            $fifthweek[$i]['id'] = date("Y-m-d",$fday+(86400 * ($i+28)));
            $fifthweek[$i]['val'] = date("d",$fday+(86400 * ($i+28)));
            if($currday_timestamp > $fday+(86400 * ($i+28))) {
                $fifthweek[$i]['flag'] = 1;    
            }
            $sixthweek[$i]['id'] = date("Y-m-d",$fday+(86400 * ($i+35)));
            $sixthweek[$i]['val'] = date("d",$fday+(86400 * ($i+35)));
            if($currday_timestamp > $fday+(86400 * ($i+35))) {
                $sixthweek[$i]['flag'] = 1;    
            }
        }

我如何解决这个闰秒问题？我需要在秒中添加一个值吗？

Answer 1

听起来这与从夏令时到标准时间的变化有关。

最简单的方法是在基准时间内添加几个小时：

$time_offset = 2 * 60 * 60; // 2 hours
$firstday = mktime(0,0,0,$month,1,$year) + $time_offset;
$day = date('Y-m-d',$firstday);
$fday = strtotime($day." last Sunday ",$firstday) + $time_offset;
$currday_timestamp = mktime(0,0,0,date('m'),date('d'),date('Y')) + $time_offset;

无法解决日历上的闰秒错误计算问题

1 个答案: