Question

鉴于$some_date等于2015-02-12 10:28:04，如何使用PHP（而非MySQL）确定它是否不比当前时间的X小？

Answer 1

strtotime和DateTime课程是我们的朋友。 $x小时为：

if(($time = strtotime($some_date)) > $time + ($x * 360)) {
    //do something its more than X hours
}

这可能不适用于夏令时间界限，所以可能：

if(($time = strtotime($some_date)) > strtotime("+$x hours", $time)) {
    //do something its more than X hours
}

Answer 2

这将确定您的日期是否距离当前日期时间不到2小时

<?php
$date = new DateTime(); //set current datetime to variable
$date->setTimezone(new DateTimeZone('America/Detroit')); //set timezone if you like
$fdate = $date->format('Y-m-d H:i:s'); //change format to year - month - day, hour, minute, seconds

$some_date = strtotime('2015-02-13 18:30:04'); // this is your datetime.  use your time or change $some_date to your variable
$new_date = strtotime($fdate) - $some_date;
$minute_date = $new_date / 60; // convert to minutes
$hour_date = $minute_date / 60; // convert to hours

print $hour_date;

if($hour_date > 2){ // change the 2 to whatever you want it to be.
    print "more than two hours";
}else{
    print "less than two hours";
}
?>

使用PHP将日期时间与当前时间进行比较

2 个答案: