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):
$dif = 1252809479 - 2009-09-12 20:57:19;
如何将我的时间戳转换成那种(unix?)格式?
再加上另一个选择……
虽然我更喜欢在这里发布的DateTime方法，但我不喜欢它显示0年等事实。
/*
* Returns a string stating how long ago this happened
*/
private function timeElapsedString($ptime){
$diff = time() - $ptime;
$calc_times = array();
$timeleft = array();
// Prepare array, depending on the output we want to get.
$calc_times[] = array('Year', 'Years', 31557600);
$calc_times[] = array('Month', 'Months', 2592000);
$calc_times[] = array('Day', 'Days', 86400);
$calc_times[] = array('Hour', 'Hours', 3600);
$calc_times[] = array('Minute', 'Minutes', 60);
$calc_times[] = array('Second', 'Seconds', 1);
foreach ($calc_times AS $timedata){
list($time_sing, $time_plur, $offset) = $timedata;
if ($diff >= $offset){
$left = floor($diff / $offset);
$diff -= ($left * $offset);
$timeleft[] = "{$left} " . ($left == 1 ? $time_sing : $time_plur);
}
}
return $timeleft ? (time() > $ptime ? null : '-') . implode(' ', $timeleft) : 0;
}
我发现结果如下:
1年2个月0天0小时53分1秒
正因为如此，我实现了一个尊重复数的函数，删除空值，并有可能缩短输出:
function since($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:
$lang = array(
'second' => 'Sekunde',
'seconds' => 'Sekunden',
'minute' => 'Minute',
'minutes' => 'Minuten',
'hour' => 'Stunde',
'hours' => 'Stunden',
'day' => 'Tag',
'days' => 'Tage',
'month' => 'Monat',
'months' => 'Monate',
'year' => 'Jahr',
'years' => 'Jahre',
'and' => 'und',
);
我知道这里有几个答案，但这是我想到的。这只处理MySQL DATETIME值，根据我回应的原始问题。数组$a需要做一些工作。我欢迎就如何改进提出意见。电话为:
Echo time_elapsed_string('2014-11-14 09:42:28');
function time_elapsed_string($ptime)
{
// Past time as MySQL DATETIME value
$ptime = strtotime($ptime);
// Current time as MySQL DATETIME value
$csqltime = date('Y-m-d H:i:s');
// Current time as Unix timestamp
$ctime = strtotime($csqltime);
// Elapsed time
$etime = $ctime - $ptime;
// If no elapsed time, return 0
if ($etime < 1){
return '0 seconds';
}
$a = array( 365 * 24 * 60 * 60 => 'year',
30 * 24 * 60 * 60 => 'month',
24 * 60 * 60 => 'day',
60 * 60 => 'hour',
60 => 'minute',
1 => 'second'
);
$a_plural = array( 'year' => 'years',
'month' => 'months',
'day' => 'days',
'hour' => 'hours',
'minute' => 'minutes',
'second' => 'seconds'
);
foreach ($a as $secs => $str){
// Divide elapsed time by seconds
$d = $etime / $secs;
if ($d >= 1){
// Round to the next lowest integer
$r = floor($d);
// Calculate time to remove from elapsed time
$rtime = $r * $secs;
// Recalculate and store elapsed time for next loop
if(($etime - $rtime) < 0){
$etime -= ($r - 1) * $secs;
}
else{
$etime -= $rtime;
}
// Create string to return
$estring = $estring . $r . ' ' . ($r > 1 ? $a_plural[$str] : $str) . ' ';
}
}
return $estring . ' ago';
}
