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):
$dif = 1252809479 - 2009-09-12 20:57:19;
如何将我的时间戳转换成那种(unix?)格式?
我发现结果如下:
1年2个月0天0小时53分1秒
正因为如此，我实现了一个尊重复数的函数，删除空值，并有可能缩短输出:
function since($timestamp, $level=6) {
global $lang;
$date = new DateTime();
$date->setTimestamp($timestamp);
$date = $date->diff(new DateTime());
// build array
$since = array_combine(array('year', 'month', 'day', 'hour', 'minute', 'second'), explode(',', $date->format('%y,%m,%d,%h,%i,%s')));
// remove empty date values
$since = array_filter($since);
// output only the first x date values
$since = array_slice($since, 0, $level);
// build string
$last_key = key(array_slice($since, -1, 1, true));
$string = '';
foreach ($since as $key => $val) {
// separator
if ($string) {
$string .= $key != $last_key ? ', ' : ' ' . $lang['and'] . ' ';
}
// set plural
$key .= $val > 1 ? 's' : '';
// add date value
$string .= $val . ' ' . $lang[ $key ];
}
return $string;
}
看起来好多了:
1年2个月53分1秒
可以选择使用$level = 2来缩短它，如下所示:
1年2个月
如果你只需要它的英文版本，就删除$lang部分，或者编辑这个翻译来满足你的需要:
$lang = array(
'second' => 'Sekunde',
'seconds' => 'Sekunden',
'minute' => 'Minute',
'minutes' => 'Minuten',
'hour' => 'Stunde',
'hours' => 'Stunden',
'day' => 'Tag',
'days' => 'Tage',
'month' => 'Monat',
'months' => 'Monate',
'year' => 'Jahr',
'years' => 'Jahre',
'and' => 'und',
);
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 $x;
}
这是我的解决方案，请检查并根据您的要求修改
function getHowLongAgo($date, $display = array('Year', 'Month', 'Day', 'Hour', 'Minute', 'Second'), $ago = '') {
date_default_timezone_set('Australia/Sydney');
$timestamp = strtotime($date);
$timestamp = (int) $timestamp;
$current_time = time();
$diff = $current_time - $timestamp;
//intervals in seconds
$intervals = array(
'year' => 31556926, 'month' => 2629744, 'week' => 604800, 'day' => 86400, 'hour' => 3600, 'minute' => 60
);
//now we just find the difference
if ($diff == 0) {
return ' Just now ';
}
if ($diff < 60) {
return $diff == 1 ? $diff . ' second ago ' : $diff . ' seconds ago ';
}
if ($diff >= 60 && $diff < $intervals['hour']) {
$diff = floor($diff / $intervals['minute']);
return $diff == 1 ? $diff . ' minute ago ' : $diff . ' minutes ago ';
}
if ($diff >= $intervals['hour'] && $diff < $intervals['day']) {
$diff = floor($diff / $intervals['hour']);
return $diff == 1 ? $diff . ' hour ago ' : $diff . ' hours ago ';
}
if ($diff >= $intervals['day'] && $diff < $intervals['week']) {
$diff = floor($diff / $intervals['day']);
return $diff == 1 ? $diff . ' day ago ' : $diff . ' days ago ';
}
if ($diff >= $intervals['week'] && $diff < $intervals['month']) {
$diff = floor($diff / $intervals['week']);
return $diff == 1 ? $diff . ' week ago ' : $diff . ' weeks ago ';
}
if ($diff >= $intervals['month'] && $diff < $intervals['year']) {
$diff = floor($diff / $intervals['month']);
return $diff == 1 ? $diff . ' month ago ' : $diff . ' months ago ';
}
if ($diff >= $intervals['year']) {
$diff = floor($diff / $intervals['year']);
return $diff == 1 ? $diff . ' year ago ' : $diff . ' years ago ';
}
}
谢谢
