我正在尝试转换格式为2009-09-12 20:57:19的时间戳，并将其转换为3分钟前用PHP。
我找到了一个有用的脚本来做这件事，但我认为它正在寻找一种不同的格式来用作时间变量。我想修改的脚本与此格式的工作是:
function _ago($tm,$rcs = 0) {
$cur_tm = time();
$dif = $cur_tm-$tm;
$pds = array('second','minute','hour','day','week','month','year','decade');
$lngh = array(1,60,3600,86400,604800,2630880,31570560,315705600);
for($v = sizeof($lngh)-1; ($v >= 0)&&(($no = $dif/$lngh[$v])<=1); $v--); if($v < 0) $v = 0; $_tm = $cur_tm-($dif%$lngh[$v]);
$no = floor($no);
if($no <> 1)
$pds[$v] .='s';
$x = sprintf("%d %s ",$no,$pds[$v]);
if(($rcs == 1)&&($v >= 1)&&(($cur_tm-$_tm) > 0))
$x .= time_ago($_tm);
return $x;
}
我认为在前几行脚本试图做的事情看起来像这样(不同的日期格式数学):
$dif = 1252809479 - 2009-09-12 20:57:19;
如何将我的时间戳转换成那种(unix?)格式?
我修改了原来的函数一点(在我看来更有用，或更符合逻辑)。
// display "X time" ago, $rcs is precision depth
function time_ago ($tm, $rcs = 0) {
$cur_tm = time();
$dif = $cur_tm - $tm;
$pds = array('second','minute','hour','day','week','month','year','decade');
$lngh = array(1,60,3600,86400,604800,2630880,31570560,315705600);
for ($v = count($lngh) - 1; ($v >= 0) && (($no = $dif / $lngh[$v]) <= 1); $v--);
if ($v < 0)
$v = 0;
$_tm = $cur_tm - ($dif % $lngh[$v]);
$no = ($rcs ? floor($no) : round($no)); // if last denomination, round
if ($no != 1)
$pds[$v] .= 's';
$x = $no . ' ' . $pds[$v];
if (($rcs > 0) && ($v >= 1))
$x .= ' ' . $this->time_ago($_tm, $rcs - 1);
return $x;
}
举个例子:
echo time_elapsed_string('2013-05-01 00:22:35');
echo time_elapsed_string('@1367367755'); # timestamp input
echo time_elapsed_string('2013-05-01 00:22:35', true);
输入可以是任何受支持的日期和时间格式。
输出:
4 months ago
4 months ago
4 months, 2 weeks, 3 days, 1 hour, 49 minutes, 15 seconds ago
功能:
function time_elapsed_string($datetime, $full = false) {
$now = new DateTime;
$ago = new DateTime($datetime);
$diff = $now->diff($ago);
$diff->w = floor($diff->d / 7);
$diff->d -= $diff->w * 7;
$string = array(
'y' => 'year',
'm' => 'month',
'w' => 'week',
'd' => 'day',
'h' => 'hour',
'i' => 'minute',
's' => 'second',
);
foreach ($string as $k => &$v) {
if ($diff->$k) {
$v = $diff->$k . ' ' . $v . ($diff->$k > 1 ? 's' : '');
} else {
unset($string[$k]);
}
}
if (!$full) $string = array_slice($string, 0, 1);
return $string ? implode(', ', $string) . ' ago' : 'just now';
}
试试这个，我从我的旧代码中找到的，它显示了正确的结果
function ago($datefrom, $dateto = -1) {
// Defaults and assume if 0 is passed in that
// its an error rather than the epoch
if ($datefrom == 0) {
return "A long time ago";
}
if ($dateto == -1) {
$dateto = time();
}
// Make the entered date into Unix timestamp from MySQL datetime field
$datefrom = strtotime($datefrom);
// Calculate the difference in seconds betweeen
// the two timestamps
$difference = $dateto - $datefrom;
// Based on the interval, determine the
// number of units between the two dates
// From this point on, you would be hard
// pushed telling the difference between
// this function and DateDiff. If the $datediff
// returned is 1, be sure to return the singular
// of the unit, e.g. 'day' rather 'days'
switch (true) {
// If difference is less than 60 seconds,
// seconds is a good interval of choice
case(strtotime('-1 min', $dateto) < $datefrom):
$datediff = $difference;
$res = ($datediff == 1) ? $datediff . ' second' : $datediff . ' seconds';
break;
// If difference is between 60 seconds and
// 60 minutes, minutes is a good interval
case(strtotime('-1 hour', $dateto) < $datefrom):
$datediff = floor($difference / 60);
$res = ($datediff == 1) ? $datediff . ' minute' : $datediff . ' minutes';
break;
// If difference is between 1 hour and 24 hours
// hours is a good interval
case(strtotime('-1 day', $dateto) < $datefrom):
$datediff = floor($difference / 60 / 60);
$res = ($datediff == 1) ? $datediff . ' hour' : $datediff . ' hours';
break;
// If difference is between 1 day and 7 days
// days is a good interval
case(strtotime('-1 week', $dateto) < $datefrom):
$day_difference = 1;
while (strtotime('-' . $day_difference . ' day', $dateto) >= $datefrom) {
$day_difference++;
}
$datediff = $day_difference;
$res = ($datediff == 1) ? 'yesterday' : $datediff . ' days';
break;
// If difference is between 1 week and 30 days
// weeks is a good interval
case(strtotime('-1 month', $dateto) < $datefrom):
$week_difference = 1;
while (strtotime('-' . $week_difference . ' week', $dateto) >= $datefrom) {
$week_difference++;
}
$datediff = $week_difference;
$res = ($datediff == 1) ? 'last week' : $datediff . ' weeks';
break;
// If difference is between 30 days and 365 days
// months is a good interval, again, the same thing
// applies, if the 29th February happens to exist
// between your 2 dates, the function will return
// the 'incorrect' value for a day
case(strtotime('-1 year', $dateto) < $datefrom):
$months_difference = 1;
while (strtotime('-' . $months_difference . ' month', $dateto) >= $datefrom) {
$months_difference++;
}
$datediff = $months_difference;
$res = ($datediff == 1) ? $datediff . ' month' : $datediff . ' months';
break;
// If difference is greater than or equal to 365
// days, return year. This will be incorrect if
// for example, you call the function on the 28th April
// 2008 passing in 29th April 2007. It will return
// 1 year ago when in actual fact (yawn!) not quite
// a year has gone by
case(strtotime('-1 year', $dateto) >= $datefrom):
$year_difference = 1;
while (strtotime('-' . $year_difference . ' year', $dateto) >= $datefrom) {
$year_difference++;
}
$datediff = $year_difference;
$res = ($datediff == 1) ? $datediff . ' year' : $datediff . ' years';
break;
}
return $res;
}
示例:echo ago('2020-06-03 00:14:21 AM');
产出:6天
