我们假设:
$time = '2010-05-17 02:49:30' // (retrieved from MySQL TIMESTAMP field)
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如何在PHP中执行以下操作:
1)检查自此时间过后是否超过一周?
2)假设(1)上的"假",找出在一周标记之前还有多少时间,四舍五入到剩余的日期和小时.
我知道这很简单,但它使用了非常具体的语法.从来没有玩过时间计算,我很感激一些指导.
谢谢!
Ann*_*rom 10
$time = strtotime('2010-05-10 02:49:30');
$one_week_ago = strtotime('-1 week');
if( $time > $one_week_ago ) {
// it's sooner than one week ago
$time_left = $time - $one_week_ago;
$days_left = floor($time_left / 86400); // 86400 = seconds per day
$hours_left = floor(($time_left - $days_left * 86400) / 3600); // 3600 = seconds per hour
echo "Still $days_left day(s), $hours_left hour(s) to go.";
}
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没有strtotime让你做这样的事情......
$timestamp = strtotime($time);
$oneweekago = strtotime("-1 week");
if($oneweekago<=$timestamp) {
// it's been less than one week
$secondsleft = $oneweekago - $timestamp;
// ...
}
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